Category: Inter 1st Year

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 6 Trigonometric Ratios up to Transformations to solve questions creatively.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Formulas

→ sin θ, cos θ, tan θ, cosec θ, sec θ and cot θ are called circular functions.

→ cosec θ, sec θ and cot θ are reciprocals to sin θ, cos θ and tan θ respectively.

→ sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ

→ sec θ + tan θ and sec θ – tan θ are mutual reciprocals.
Similarly cosec θ + cot θ and cosec θ – cot θ are also mutual reciprocals.

→ |sin θ| ≤ 1, |cosec θ| ≥ 1, |cos θ| ≤ 1 and |sec θ| ≥ 1

• sin 0° = o = cos 90°
• sin 15° = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\) = cos 75°
• sin 18° = \(\frac{\sqrt{5}-1}{4}\) = cos 72°
• sin 30° = \(\frac{1}{2}\) = cos 60°
• sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\) = cos 54°
• sin 45° = \(\frac{1}{\sqrt{2}}\) = cos 45°
• sin 54° = \(\frac{\sqrt{5}+1}{4}\) = cos 36°
• sin 60° = \(\frac{\sqrt{3}}{2}\) = cos 30°
• sin 72° = \(\frac{\sqrt{10+2 \sqrt{5}}}{4}\) = cos 18°
• sin 75° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\) = cos 15°
• sin 90° = 1 = cos 0°

Also, tan 15° = 2 – √3, tan 75° = 1 + √3
Sign to the determined by “ALL SILVER CUPS” rules

• sin (90° – θ) = cos θ; sin (90° + θ) = cos θ
• cos (90° – θ) – sin θ; cos (90° + θ) = -sin θ
• sin (180° – θ) = sin θ; sin (1800 + θ)= -sin θ
• cos (180° – θ) = -cos θ; cos (180° + θ) = -cos θ
• sin (270° – θ) = -cos θ; sin (270° + θ) = -cos θ
• cos (270° – θ) = -sin θ; cos (270° + θ) = sin θ
  sin (360° – θ) = -sin θ; sin (360° + θ) = sin θ
• cos (360° – θ) = cos θ; cos (360° + θ) = cos θ When ‘n’ is a +ve integer,
• sin (n. 360° – θ) = -sin θ; sin (n. 360° + θ) = +sin θ
• cos (n. 360° – θ) = cos θ; cos (n. 360° + θ) = cos θ
• sin (-θ) = – sin θ, cos (-θ) = cos 0; tan (-θ) = -tan θ.

→ For 0°, 180°, 360° ……………….. (multplies of π), there is no change in the ratios.

→ For 90°, 270°, 450°…………. (odd multiplies of \(\frac{\pi}{2}\) we get change in the ratio.

• For sin we get cos
• For tan we get cot
• For sec we get cosec
• For cos we get sin
• For cot we get tan
• For cot we get sin

→ Any non-constant function f : R → R is said to be “Periodic”, if there exists a real number p(≠ 0) such that f(x + p) = f(x) for each x ∈ R. The least positive value of ‘p’ with this property is called the “Period” of ‘f’.

→ If ‘f’ is a periodic function with period ‘P’ then

• – f is also a periodic function with period ‘P.
• f(x – p) = f(x), f(x + pn) = f(x) ∀ n ∈ Z, x ∈ R.

→ If f(x) is a periodic function with period ‘P, then f(ax + b) is also a periodic function with period \(\frac{P}{|a|}\)

→ If y = f(x), y = g(x) are periodic functions with l, m as the periods respectively, then a, b ∈ R the function h(x) = af(x) + bg(x) is a period function and L.C.M. of {l, m}
(if exist) is a period of h.

→ \(\frac{2 \pi}{k}\) is the period of sin x, cosec x, cos x and sec x.

→ \(\frac{\pi}{k}\) is the period of tan x and cot x.

→ Range of a sin x + b cos x is \(\left[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}\right]\)

→ Range of a sin x + b cos x + c is [c – \(\sqrt{a^{2}+b^{2}}\), c + \(\sqrt{a^{2}+b^{2}}\)]

→ sin x and cos x are continuous on ‘R’

→ tan x is. discontinuous at x = (2n + 1).\(\frac{\pi}{2}\),n ∈ Z.

→ cot x is discontinuous at x = nπ, n ∈ Z.

→ sec x is discontinuous at x = (2n + 1).\(\frac{\pi}{2}\), n ∈ Z.

→ cosecx is discontinuous at x = nπ, n ∈ Z.

→ sin (A ± B) = sin A cos B ± cos A sin B

→ cos (A ± B) = cos A cos B + sin A sin B

→ tan (A ± B) = \(\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

→ cot (A ± B) = \(\frac{\cot A \cot B \mp 1}{\cot B \pm \cot A}\)

→ sin (A + B). sin (A-B) = sin²A – sin²B = cos²B – cos²A

→ cos (A + B). cos (A – B) = cos²A – sin²B = cos²B – sin²A

→ sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C

→ cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C

→ tan (A + B + C) = \(\frac{\sum \tan A-\pi \tan A}{1-\sum \tan A \tan B}\)

→ sin 2A = 2 sin A cos A, sin A = 2 sin \(\frac{A}{2}\) cos \(\frac{A}{2}\)

→ cos 2A = cos²A – sin²A = 1 – 2 sin² A = 2 cos²A – 1

→ cos A = cos²\(\frac{A}{2}\) sin²\(\frac{A}{2}\) = 1 – 2 sin²\(\frac{A}{2}\) = 2 cos² \(\frac{A}{2}\) – 1

→ tan 2A = \(\frac{2 \tan A}{1-\tan ^{2} A}\), tan A = \(\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}\)(\(\frac{A}{2}\), A are not odd multiples of \(\frac{\pi}{2}\))

→ cot 2A = \(\frac{\cot ^{2} A-1}{2 \cot A}\), cot A = \(\frac{\cot ^{2} \frac{A}{2}-1}{2 \cot \frac{A}{2}}\); (A is not an integral multiple of π)

→ sin 2A = \(\frac{2 \tan A}{1+\tan ^{2} A}\), sin A = \(\frac{2 \tan \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}\); (\(\frac{A}{2}\), is not an odd multiple of \(\frac{\pi}{2}\))

→ cos 2A = \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\), cos A = \(\frac{1-\tan ^{2} \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}\)

→ sin 3A = 3sin A – 4 sin³A

→ cos 3A = 4cos³A – 3 cos A

→ tan 3A = \(\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}\)

→ cot 3A = \(\frac{3 \cot A-\cot ^{3} A}{1-3 \cot ^{2} A}\)

→ sin(A + B) + sin (A – B) = 2sin A cos B

→ sin(A + B) – sin(A – B) = 2cos A sin B

→ cos (A + B) + cos (A – B) = 2cos A cos B

→ cos(A + B) – cos(A – B) = -2sin A sin B

→ sin C + sin D = 2sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

→ sin C – sin D = 2cos \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)

→ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

→ cos C – cos D = -2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)

→ For any A ∈ R
(a) sin A = ±\(\sqrt{\frac{1-\cos 2 A}{2}}\)
(b) cos A = ±\(\sqrt{\frac{1+\cos 2 A}{2}}\)
(c) IfA ¡s not an odd multiple of \(\frac{\pi}{2}\), then tan A = ±\(\sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}\)

→ sin \(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}\)

→ cos \(\frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}\)

→ If A is not an odd multiple of n, then tan \(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\)

→ If a ray \(\overrightarrow{\mathrm{OP}}\) makes an angle θ with the positive direction of X-axis then

• Sin θ = \(\frac{\mathrm{y}}{\mathrm{r}}\)
• cos θ = \(\frac{\mathrm{x}}{\mathrm{r}}\)
• tan θ = \(\frac{\mathrm{y}}{\mathrm{x}}\) (x ≠ 0)
• cot θ = \(\frac{\mathrm{x}}{\mathrm{y}}\) (y ≠ 0)
• sec θ = \(\frac{\mathrm{r}}{\mathrm{x}}\)(x ≠ 0)
• cosec θ = \(\frac{\mathrm{r}}{\mathrm{y}}\)(y ≠ 0)

Relations :

• sin θ cosec θ = 1
• cos θ sec θ = 1
• tan θ cot θ = 1
• sin² θ + cos² θ = 1
• 1 + tan²θ = sec² θ → (sec θ + tan θ)(sec θ – tan θ) = 1
  → sec θ + tan θ = \(\frac{1}{\sec \theta-\tan \theta}\) = 1
• 1 + cot²θ = cosec²θ → (cosec θ + cot θ) (cosec θ – cot θ) = 1
• sec²θ + cosec²θ = sec²θ. cosec²θ
• tan²θ – sin²θ = tan²θ . sin²θ;
  cot²θ – cos²θ = cot²θ. cos²θ
• sin²θ + cos⁴θ = 1 – sin²θ cos²θ
  = sin⁴θ + cos²θ
• sin⁴θ + cos⁴θ = 1 – 2sin²θ cos²θ
• sin⁶θ + cos⁶ θ = 1 – 3sin²θ cos²θ
• sin²x + cosec²x ≥ 2
• cos²x + sec² x ≥ 2
• tan² x + cot² x ≥ 2.

Values of trigonometric ratios of certain angles

Signs of Trigonometric ratios :
If lies in I, II, III, IV quadrants then the signs of trigonometric ratios are as follows.

Note :

• 0°, 90°, 180°, 270°. 360°, 450°, etc. are called quadrant angles.
• With “ALL SILVER TEA CUPS” symbol we can remember the signs of trigonometric ratios.

Coterminal angles :
If two angles differ by an integral multiples of 360o then two angles are called coterminal angles.
Thus 30°, 390°, 750°, 330° etc., are coterminal angles.

Complementary Angles :
Two Angles A, B are said to complementary ⇒ A + B = 90°

Supplementary angles :
Two angles A, B are said to be supplementary ⇒ A + B = 180°.

→ sin C + sin D = 2sin\(\frac{\mathrm{C}+\mathrm{D}}{2}\). cos \(\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ sin C – sin D = 2cos\( . sin [latex]\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ cos C + cos D = 2cos \(\frac{\mathrm{C}+\mathrm{D}}{2}\). cos \(\frac{\mathrm{C}-\mathrm{D}}{2}\)

→ cos C – cos D = 2sin\(\frac{\mathrm{C}+\mathrm{D}}{2}\). sin \(\frac{\mathrm{D}-\mathrm{C}}{2}\)

→ 2sin A cos B = sin(A + B) + sin(A – B)

→ 2cos A sin B = sin(A + B) – sin(A – B)

→ 2cos A cos B = cos(A + B) + cos(A – B)

→ 2sin A sin B = cos(A – B) – cos(A + B)
(or)
cos(A – B) – cos(A + B) = 2 sin A sin B.

→ \(\frac{\sin A+\sin B}{\sin A-\sin B}\) = tan\(\left(\frac{A+B}{2}\right)\).

→ If sin A + sin B = x, and cos A + cos B = y. Then

• tan\(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\frac{\mathrm{x}}{\mathrm{y}}\)
• sin(A + B) = \(\frac{2 x y}{y^{2}+x^{2}}\)
• cos (A + B) = \(\frac{y^{2}-x^{2}}{y^{2}+x^{2}}\)
• tan(A + B) = \(\frac{2 x y}{y^{2}-x^{2}}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 5 Products of Vectors to solve questions creatively.

Intermediate 1st Year Maths 1A Products of Vectors Formulas

Scalar or Dot Product of Two Vectors:
The scalar or dot product of two non – zero vectors \(\bar{a}\) and \(\bar{b}\), denoted by \(\bar{a} \cdot \bar{b}\) is defined as \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos \((\bar{a}, \bar{b})\). This is a scalar, either \(\bar{a}\) = 0 (or) \(\bar{b}\) = 0, then we define \(\bar{a} \cdot \bar{b}\) = 0. If we write \((\bar{a}, \bar{b})\) = 0, then \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}|\) cos θ, if a ≠ 0, b ≠ 0, since 0 ≤ (a, b) = θ ≤ 7 80°, we get

• 0 ≤ θ < 90° ⇒ \(\bar{a}\). b > 0.
• θ = 90° ⇒ \(\bar{a} \cdot \bar{b}\) = 0 and the vectors \(\bar{a}\) and \(\bar{b}\) are perpendicular.
• 90° < θ ≤ 180° ⇒ \(\bar{a} \cdot \bar{b}\) < 0
• \(\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}\)
• a̅ (b̅ + c̅) = a̅ .b̅ + a̅ .c̅
• If a̅, b̅ are parallel, a̅.b̅ = ± |a̅ | |b̅ |.
• If l, m ∈ R, (la̅).(mb̅) = lm(a̅. b̅)
• Projection of b̅ on a̅ (or) length of the projection a̅ = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
• Orthogonal projection of b̅ on a̅ = \(\frac{(\bar{a} \cdot \bar{b})_{\bar{a}}}{|\bar{a}|^{2}}\); a̅ ≠ 0
  or
  The projection vector b̅ on a̅ = \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\) a̅ and it is magnitude = \(\frac{|\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}|}\)
• The component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^{2}}\right)\)a̅

Component vector of a̅ along b̅ = \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\), component vector of a̅ perpendicular to b̅ = a̅ – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^{2}}\)

Orthogonal unit vectors :
Ifi, j, k are orthogonal unit vector triad in a right handed system, then

• i̅ .j̅ = j̅.k̅ = k̅.i̅ = 0
• i̅ .i̅ = j̅.j̅ = k̅.k̅ = 1
• If r is any vector, r̅ = (r̅.i̅)i̅ +(r̅.j̅)j̅ ≠ (r̅.k̅)k̅

Some identities :
If a̅, b̅, c̅ are three vectors, then

• (a̅ + b̅)² = |a̅|² + |b̅|² + 2(a̅ . b̅)
• (a̅ – b̅)² = |a̅|² + |b̅|² – 2(a̅.b̅)
• (a̅ + b̅)² + (a̅ – b̅)² = 2(|a̅|² + |b̅|²)
• (a̅ + b̅)² – (a̅ – b̅)² = 4(a̅. b̅)
• (a̅ + b̅). (a̅ – b̅) = |a̅|² – |b̅|²
• (a̅ + b̅ + c̅)² = |a̅|² + |b̅|² + |c̅|² + 2(a̅ . b̅) + 2(b̅ . c̅) + 2(c̅ .a̅)

→ If a̅ = a₁ i̅ +a₂j̅ + a₃k̅ and b̅ = b₁i̅ + b₂j̅ + b₃k̅, then
a̅.b̅ = a₁b₁ + a₂b₂ + a₃b₃
a̅ is perpendicular to b̅
⇔ a₁b₁ + a₂b₂ + a₃b₃ = 0

→ |a̅| = \(\), |b̅| = \(\)

→ If (a̅, b̅) = then cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) and sin θ = \(\sqrt{\frac{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}{\left(\sum a_{1}^{2}\right)\left(\sum b_{1}^{2}\right)}}\)

→ a̅ is parallel to b̅ ⇔ a₁: b₁ = a₂ : b₂ = a₃: b₃

→ a̅.a̅ >0; |a̅.b̅| < |a̅| |b̅|
|a̅ + b̅| ≤ |a̅| + |b̅|; |a̅ – b̅| ≤ |a̅| + |b̅| ;
|a̅ – b̅| ≥ |a̅| – |b̅|

Vector equations of a plane :

• The equation of the plane, whose perpendicular distance from the origin is p and whose unit normal drawn from the origin towards the plane is h is n̂ is r̅.n̂ = p.
• Equation of a plane passing through the origin and perpendicular to the unit vector n̅, is r̅.n̅ = 0
• Vector equation of a plane passing through a point A with position vector a and perpendicular to a vector n̅ is (r – a̅). n̅ = 0.

Perpendicular distance from the origin to the plane (r̅ – a̅).h = 0 is a̅. n̅ . where ‘a̅’ is the position vector of A in the plane and ‘n̅’ is a unit vector perpendicular to the plane.

Angle between two planes :
If π₁ and π₂2 be two planes and \(\bar{M}_{1}, \bar{M}_{2}\) are normals drawn to them, we define the angle between M₁ and M₂ as the angle between π₁ and π₂. If the angle between \(\) and \(\) is θ, the angle between the given planes θ = cos^-1\(\left[\frac{\bar{M}_{1} \cdot \bar{M}_{2}}{\left|\bar{M}_{1}\right|\left|\bar{M}_{2}\right|}\right]\)

Work done by a constant force F:

• If a constant force F̅ acting on a particle displaces it from a position ‘A’ to the position B, then the work done ‘W by this constant force T is the dot product of the vectors
  representing the force F̅ and displacement \(\overline{A B}\), i.e., W = F̅.\(\overline{A B}\).
• If F is the resultant of the forces F̅₁, F̅₂, ……………….F̅_n, then workdone in displacing the particle from A to B is
  \(\bar{W}=\bar{F}_{1} \cdot \overline{A B}+\bar{F}_{2} \cdot \overline{A B}+\ldots \ldots+F_{n} \cdot \overline{A B}\)

Cross Product or Vector Product of two vectors :
The vector product or cross product of two non-parallel non – zero vectors ‘a̅’ and ‘b̅’ is defined as a̅ × b̅ = |a̅||b̅| sin θ n̂, where ‘ n̂’ is a unit vector perpendicular to the plane containing ‘a̅’ and ‘b̅’ such that a̅, b̅ and ‘n̂’ form a vector triad in the right handed system and (a̅, b̅) = θ, this is a vector. If either of a̅, b̅ is a zero vector or ‘a̅’ is parallel to ‘b̅’, we define a̅ × b̅ = 0.

Some important results on vector product:

• |a̅ × b̅| = |a̅||b̅|sinθ ≤ |a̅||b̅| ;
• |a̅ × b̅| = |b̅ × a̅|
• a̅ × b̅ = -(b̅ × a̅):
• -a̅ × -b̅ = a̅ × b̅
• (-a̅) × b̅ = a̅ × (-b̅) – (a̅ × b̅)
• la̅ × mb̅ = lm(a̅ × b̅) ;
• a̅ × (b̅ + c̅) = a̅ × b̅ + a̅ × c̅
• a̅ ≠ 0, b̅ ≠ 0 and a̅ × b̅ = 0 ⇔ ‘a̅’ and ‘b̅’ are parallel vectors.
• a̅, b̅ , c̅ are non-zero vectors and a̅ × c̅ = b̅ × c̅ ⇒ either a̅ = b̅ or a̅ – b̅ is parallel to c̅.

Vector product among i. i and k:
If i̅, j̅ and k̅ are orthogonal unit vectors triad in the right handed system then

• i̅ × j̅ = j̅ × j = k̅ × k̅ = 0
• i̅ × j̅ = k̅ =-j̅ × i̅ ; j̅ × k̅ = k̅ × j̅ = i̅ ; k̅ × i̅ = -i̅ × k̅ = j̅
• If a̅ = a₁ i̅ + a₂ j + a₃k ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅, then
  a̅ × b̅ = a₂b₃ – a₃b₂)i̅ + (a₃b₁ – a₁b₃)j̅ + (a₁b₂ – a₂b₁)k̅

This may be represented in the form of a determinants as a̅ × b̅ = \(\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

• Unit vectors perpendicular to both ‘a̅’and ‘b̅’ are ± \(\frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}\)
• If a̅ = a₁i̅ + a₂j̅ + a₃k̅ ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅ and (a̅, b̅) = θ, then
  sin θ = \(\frac{\sqrt{\sum\left(a_{2} b_{3}-a_{3} b_{2}\right)^{2}}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\) cos θ = \(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{\sum a_{1}^{2}} \sqrt{\sum b_{1}^{2}}}\)

Vector areas:

• If \(\overline{A B}=\bar{c}\) and \(\overline{A C}=\bar{b}\) are two adjacent sides of a triangle ABC, then vector area of ΔABC = \(\frac{1}{2}\)(c̅ × b̅) and the area of the ΔABC = \(\frac{1}{2}\)|c̅ × b̅| $q. units.
• If a̅, b̅, c̅ are the position vectors of A, B, C respectively then the vector area of
  ΔABC = \(\frac{1}{2}\)[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)]
  Area of ΔABC = \(\frac{1}{2}\)|(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)|sq. units.
• If \(\) and \(\) are the diagonals of a parallelogram ABCD, then the vector area of the parallelogram = \(\frac{1}{2}\)|a̅ × b̅| and area = \(\frac{1}{2}\)|a̅ × b̅|sq. units.
• If AB = a̅ and AD = b̅ are two adjacent sides of a parallelogram ABCD, then its vector area = a̅ × b̅ and area = |a̅ × b̅| sq. units.
• Vector area of the quadrilateral ABCD = \(\frac{1}{2}\)\((A C \times B D)\) and area of the quadrilateral ABCD = \(\frac{1}{2}\)\(|\overline{A C} \times \overline{B D}|\)sq. units.

Some useful formulas :

• If a̅, b̅ are two non-zero and non-parallel vectors then
  (a̅ × b̅)² = a²b² – (a̅.b̅)² = \(\left|\begin{array}{cc}
  a \cdot \bar{a} & \bar{a} \cdot \bar{b} \\
  \bar{a} \cdot b & b \cdot \bar{b}
  \end{array}\right|\)
• For any vector a̅,(a̅ × i̅)² + (a̅ × j̅)² + (a̅ × k̅)² = 2|a|²
• If a̅, b̅, c̅ are the position vectors of the points A, B, C respectively, then the perpendicular distance from c to the line AB is \(\frac{|\overline{A C} \times \overline{A B}|}{|\overline{A B}|}=\frac{|(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})+(\bar{a} \times \bar{b})|}{|b-\bar{a}|}\)

Moment of a force :
Let 0 be the point of reference (origin) and \(\overline{o p}=\bar{r}\) be the position vector of a point p on the line of action of a force F̅. Then the moment of the force F about 0 is given by r̅ × F̅.

Scalar triple product:
Let a̅, b̅, c̅ he three vectors. We call (a̅ × b̅). c̅ the scalar product of a̅, b and c. This is a scalar (real number). It is written as [a̅ b̅ c̅]

• If (a̅ × b̅). c̅ = 0, then one or more of the vectors a̅, b̅ and c̅ should be zero vectors. If a ≠ 0, b ≠ 0, c ≠ 0, then c is perpendicular to a̅ × b̅. Hence the vector c̅ lies on the plane determined by a̅ and b̅. Hence a̅, b̅ and c̅ are coplanar.
• If in a scalar triple product, any two vectors are parallel (equal), then the scalar triple product is zero i.e., [a̅ a̅ b̅] = [a̅ b̅ b̅] = [c̅ b̅ c̅] = 0.
• In a scalar triple product remains unaltered if the vectors are permutted cyclically i.e., [a̅ b̅ c̅] = [b̅ c̅ a̅] = [c̅ a̅ b̅].
  However [a̅ b̅ c̅] = -[b̅ a̅ c̅] = -[c̅ b̅ a̅] = -[a̅ c̅ b̅].
• In a scalar triple product, the dot and cross are interchangeable i.e., a̅.b̅ × c̅ = a̅ × b̅.c̅

→ If i̅ , j̅ , k̅ are orthogonal unit vector triad in the right handed system, then

• [i̅ j̅ k̅ ] = [j̅ k̅ i̅ ] = [k̅ i̅ j̅ ] = 1
• [i̅ k̅ j̅ ] = [j̅ i̅ k̅ ] = [k̅ j̅ i̅ ] = -1
• If a̅ = a₁ i̅ + a₂j̅ + a₃ k̅ ; b̅ = b₁i̅ + b₂ j̅ + b₃k̅ and c̅ = c₁i̅ + c₂ j̅ + c₃k̅ then [a̅ b̅ c̅] = \(\left|\begin{array}{lll}
  a_{1} & a_{2} & a_{3} \\
  b_{1} & b_{2} & b_{3} \\
  c_{1} & c_{2} & c_{3}
  \end{array}\right|\)

→ A necessary and sufficient condition that three non-parallel (non-collinear) and non-zero vectors a, b and c to be coplanar is [a̅ b̅ c̅] = 0. If [a̅ b̅ c̅] ≠ 0, then the three vectors are non-coplanar.

→ If a̅, b̅, c̅ are three non-zero, non-coplanar vectors and V is the volume of the parallelopiped with co-terminus edges a̅, b̅ and c̅, then v = |(a × b). c|. = |[a b c]|

→ The volume of the parallelopiped formed with A, B, C, D as vertices is \(|[A B A C A D]|\) cubic units.

→ If a̅, b̅, c̅ represent the co-terminus edges of a tetrahedron, then its volume = \(\frac{1}{6}\)[a̅, b̅, c̅] cubic units.

→ If A(x₁ y₁ z₁], B(x₂, y₂, z₂] C(x₃ y₃ z₃) and D(x₄, y₄ z₄] are the vertices of a tetrahedron = \(\frac{1}{6}\)\(|[A B A C A D]|\)

• Vector equation of a plane containing three non-collinear points a̅, b̅, c̅ is r̅ .[(b̅ × c̅) + (c̅ × a̅) + (a̅ × b̅)] = [a̅ b̅ c̅]
• A unit vector perpendicular to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{(\bar{a} \times b)+(b \times \bar{c})+(\bar{c} \times \bar{a})}{|(\bar{a} \times \bar{b})+(b \times \bar{c})+(\bar{c} \times a)|}\)
• Length of the perpendicular from the origin to the plane containing three non-collinear points a̅, b̅, c̅ is \(\frac{|[\bar{a} b c]|}{|(\bar{a} \times \bar{b})+(\bar{b} \times \bar{c})+(\bar{c} \times \bar{a})|}\)

→ Vector equation of the plane passing through three non-collinear points a̅, b̅ and c̅ is [r̅ – a̅ b̅ – a̅ c̅ – a̅] = 0

→ Vector equation of the plane passing through a given point a̅ and parallel to the vectors b̅ and c̅ is [r b̅ c̅] = [a̅ b̅ c̅]

→ Vector equation of the line passing through the point a̅ and parallel to the vector b̅ is (r – a̅) × b̅ = 0

→ Distance of the point p(c) from a line joining the points A(a) and B(b) = |(c̅ – a̅) × b̅|
(25) i) Equation of the plane passing through the point p(x₁, y₁, z₁) and perpendicular to the vector ai̅ + bj̅ + ck̅ is a (x – x₁) + b(y – y₁) + c(z – z₁) = 0.

→ The equation of the plane passing through the points (x₁ y₁ z₁), (x₂, y₂, z₂) and
(x₃, y₃, z₃) is \(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0

Skew lines:
Two lines l and m are called skew lines if there is no plane passing through these lines.

Shortest distance between the skew lines:
Shortest distance between the skew lines r̅ = a̅ + tb̅ and r̅ = c̅ + sd̅ is

Vector triple product:
If a̅, b̅ and c̅ are three vectors, products of the type (a̅ × b̅) × c̅, a̅ × (b̅ × c̅) from the vector triple products. From this definition.

• If any one of a̅, b̅ and c̅ is a zero vector, a̅ × (b̅ × c̅) or (a̅ × b̅) c̅ = 0
• If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and a̅ is parallel to b, then (a̅ × b̅) × c̅ = 0
• If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0 and c̅ is perpendicular to the plane of a and b, then (a̅ × b̅) × c̅ = 0.

→ If a̅ ≠ 0, b̅ ≠ 0, c̅ ≠ 0, a̅ and b̅ are non-parallel vectors and c̅ is not perpendicular to the plane passing through a̅ and b, then
(a̅ × b̅) × c̅ = (a̅.c̅)b̅ – (b̅.c)a̅
a̅ × (b̅ × c̅) = (a̅.c̅)b̅ – (a.b̅)c̅

• In general, vector triple product of three vectors need not satisfy the associative law. i.e., (a̅ × b) × c ≠ a̅ × (b × c)
  For any three vectors a̅, b̅ and c
• a̅ × (b̅ × c̅) + b̅ × (c̅ × a̅) + c̅ × (a̅ × b̅) = 0
• [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]²

Scalar product of four vectors :
Scalar product of a̅, b̅, c̅ and d̅ is () = (a̅. c̅) (b̅.d̅) – (a̅. d̅) (b̅ .c̅) = \(\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

Vector product of four vectors:
If a̅, b̅, c̅ and d̅ are four vectors,
(a̅ × b̅) × (c̅ × d̅) = [a̅ c̅ d̅] b̅ – [b̅ c̅ d̅]a̅ – [a b̅ d̅]c̅ – [a̅ b̅ c̅]d̅

Some important results:

• [a̅ + b̅ b̅ + c̅ c̅ + a̅] = 2[a̅ b̅ c̅]
• i̅ × (j̅ × k̅) + i̅ × (k̅ × i̅) + k̅ × (i̅ × j̅) = 0
• [a̅ × b̅ b̅ × c̅ c̅ × a̅] = [a̅ b̅ c̅]²
  [a̅ b̅ c̅] [l̅ m̅ n̅] = \(\)
• If a̅, b̅, c̅ be such that a is perpendicular to (b̅ + c̅), bis perpendicular to (c̅ + a̅), c̅ is perpendicular to (a̅ + b̅), then |a̅ + b̅ + c̅| = \(\sqrt{a^{2}+b^{2}+c^{2}}\)
• If a line makes angles α, β, γ and δ with the diagonals of a cube, then
  cos²α + cos²β + cos²γ + cos²δ = \(\frac{4}{3}\)
• i̅ × (a̅ × i̅) + j̅ × (a̅ × j̅) + k̅ × (a̅ × k̅) – 2a̅

→ Equation of the sphere with centre at c and radius ‘a’ is r² – 2r̅. c̅ + c² = a²

Scalar Product
Def: Let \(\vec{a}, \vec{b}\) be two vectors dot product (or) scalar product (or) direct product (or) inner product denoted by \(\vec{a}, \vec{b}\). Which is defined as \(|\vec{a}||\vec{b}|\)cos θ where cos θ = \((\vec{a}, \vec{b})\)
* The product \(\vec{a}, \vec{b}\) is zero when \(|\vec{a}|\) = 0 (or) \(|\vec{b}|\) = 0 (or) θ = 90°.

Sign of the scalar product :
Let \(\vec{a}, \vec{b}\) are two non-zero vectors

• If θ is acute then \(\vec{a}.\vec{b}\)> 0 (i.e 0 < θ < 90°).
• If θ is obtuse then \(\vec{a}.\vec{b}\) < 0 (i.e 90° < θ < 180°).
• If θ = 90° then\(\vec{a} \cdot \vec{b}\) = o.
• If θ = 0° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)
• If θ = 180° then \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\)

Note:

• The dot product of two vectors is always scalar.
• \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) i.e dot product of two vectors is commutative.
• If \(\vec{a} \cdot \vec{b}\) are two vectors then \(\vec{a} \cdot(-\vec{b})=(-\vec{a}) \cdot \vec{b}=-(\vec{a} \cdot \vec{b})\)
• \((-\vec{a}) \cdot(-\bar{b})=\vec{a} \cdot \vec{b}\)
• If l,m are two scalars and \(\vec{a} \cdot \vec{b}\) are two vectors then \((l \bar{a}) \cdot(m \bar{b})={lm}(\vec{a} \cdot \vec{b})\)
• If \(\vec{a}\) and \(\vec{b}\) are two vectors then \(\vec{a} \cdot \vec{b}=\pm|\vec{a}||\vec{b}|\)
• If \(\vec{a}\) is a vector then \(\vec{a} \cdot \vec{a}=|\vec{a}|^{2}\)
• If \(\vec{a}\) is a vector \(\vec{a}\).\(\vec{a}\) is denoted by \(\overline{(a)^{2}}\) hence \(\overline{(a)^{2}}=|\vec{a}|^{2}\)

Components and orthogonal projection:
Def: Let \(\vec{a}=\overline{O A} \quad \vec{b}=\overline{O B}\) be two non zero vectors let the plane passing through B and perpendicular to a intersect \(\overline{O A}\) ln M.

• If \((\vec{a}, \vec{b})\) is acute then OM is called component of \(\vec{b}\) on \(\vec{a}\).
• If \((\vec{a}, \vec{b})\) is obtuse then -(OM) is called the component of \(\vec{b}\) on \(\vec{a}\).
• The vector \(\overline{O M}\) is called component vector of \(\vec{b}\) on \(\vec{a}\).

Def: Let \(\vec{a}=\overline{O A}\); \(\vec{b}=\overrightarrow{P Q}\) be two vectors let the planes passing through P, Q and perpendicular to a intersect \([latex]\)[/latex] in L, M respectively then \(\overline{L M}\) is called orthogonal projection of \(\vec{b}\) on \(\vec{a}\)

Note:

• The orthogonal projection of a vector b on a is equal tb component vector of b on a .
• Component of a vector \(\vec{b}\) on \(\vec{a}\) is also called projection of \(\vec{b}\) on \(\vec{a}\)
• If A< B, C, D are four points in the space then the component of \(\overline{A B}\) on \(\overline{C D}\) is same as the projection of \(\overline{A B}\) on the ray \(\overline{C D}\).

→ If \(\vec{a}, \vec{b}\) be two vectors (\(\vec{a} \neq \vec{o}\)) then

• The component of \(\vec{b}\) on \(\vec{a}\) is \(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}\)
• The orthogonal projection of \(\vec{b}\) on \(\vec{a}\) is \(\frac{(\vec{b} \cdot \vec{a}) \vec{a}}{|\vec{a}|^{2}}\)

→ If \(\vec{i}, \vec{j}, \vec{k}\) form a right handed system of Ortho normal triad then

• \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}\) = 1
• \(\vec{i} \cdot \vec{j}=\vec{j} \cdot \vec{i}=0 ; \vec{j} \cdot \vec{k}=\vec{k} \cdot \vec{j}=0 ; \vec{k} \cdot \vec{i}=\vec{i} \cdot \vec{k}=0\)

→ If \(\vec{a}=a_{1} \vec{i}+a_{2} \vec{j}+a_{3} \vec{k}\) \(\vec{b}=b_{1} \vec{i}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \cdot \vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\)

→ If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then

→ If \(\vec{r}\) is vector then \(\vec{r}=(\vec{r} \cdot \vec{i}) i+(\vec{r}+\vec{j}) \vec{j}+(\vec{r} \cdot \vec{k}) \vec{k}\)

Angle between the planes:
The angle between the planes is defined as the angle between the normals to the planes drawn from any point in the space.

Sphere:
The vector equation of a sphere with centre C having position vector \(\vec{c}\) and radius a is \((\vec{r}-\vec{c})^{2}\) = a² i.e. \(\vec{r}^{2}-2 \vec{r} \cdot \vec{c}+c^{2}\) = a²

The vector equation of a sphere with A(a) and B(b) as the end points of a diameter is \((\vec{r}-\vec{a}) \cdot(\vec{r}-\vec{b})\) = 0 (or) \((\vec{r})^{2}-\vec{r} \cdot(\vec{a}+\vec{b})+\vec{a} \cdot \vec{b}\) = 0

Work done by a force :
If a force \(\vec{F}\) acting on a particle displaces it from a position A to the position B then work done W by this force \(\vec{F}\) is \(\vec{F} \cdot \overline{A B}\)

• The vector equation of the plane which is at a distance of p from the origin along the unit vector \(\vec{n}\) is \(\vec{r} \cdot \vec{n}\) = p.
• The vector equation of the plane passing through the origin and perpendicular to the vector m is r.m =0
• The Cartesian equation of the plane which is at a distance of p from the origin along the unit vector n = li + mj + nk of the plane is n = lx + my + nz
• The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and perpendicular to the vector \(\vec{m}\) is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.
• The vector equation of the plane passing through the point a having position vector \(\vec{a}\) and parallel to the plane r.m=q is \((\vec{r}-\vec{a}) \cdot \vec{m}\) = 0.

Cross( Vector) Product of Vectors:
Let \(\vec{a}, \vec{b}\) be two vectors. The cross product or vector product or skew product of vectors \(\vec{a}, \vec{b}\) is denoted by \(\vec{a} \times \vec{b}\) and is defined as follows

• If \(\vec{a}\) = 0 or \(\vec{b}\) = 0 or \(\vec{a}, \vec{b}\) are parallel then \(\vec{a} \times \vec{b}\) = 0
• If \(\vec{a}\) ≠ 0, \(\vec{b}\) ≠ 0, \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}|(\sin \theta) \vec{n}\) where \(\vec{n}\) is a unit vector perpendicular to a and b so that a, b, n form a right handed system.

Note:

• \(\vec{a} \times \vec{b}\) is a vector
• If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a} \times \vec{b}\) is perpendicular to both a and b
• If \(\vec{a}, \vec{b}\) are not parallel then \(\vec{a}, \vec{b}\) , \(\vec{a} \times \vec{b}\) form a right handed system .
• For any vector \(\vec{a}\) \(\vec{a} \times \vec{b}\) = o

2. If \(\vec{a}, \vec{b}\) are two vectors \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\) this is called “anti commutative law”

3. If \(\vec{a}, \vec{b}\) are two vectors then \(\vec{a} \times(-\vec{b})=(-\vec{a}) \times \vec{b}=-(\vec{a} \times \vec{b})\)

4. If \(\vec{a}, \vec{b}\) are two vectors then \((-\vec{a}) \times(-\vec{b})=\vec{a} \times \vec{b}\)

5. If \(\vec{a}, \vec{b}\) are two vectors l,m are two scalars then (la) x (mb) = lm(a x b)

6. If \(\vec{a}, \vec{b}, \vec{b}\) are three vectos, then

• \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)
• \((\vec{b}+\vec{c}) \times \vec{a}=\vec{b} \times \vec{a}+\vec{c} \times \vec{a}\)

7. If \(\vec{l}, \vec{l}, \vec{k}\) from a right handed system of orthonormal triad then

• \(\vec{l} \times \vec{l}=\vec{j} \times \vec{j}=\vec{k} \times \vec{k}=\vec{o}\)
• \(\vec{i} \times \vec{j}=\vec{k}=-\vec{j} \times \vec{l} ; \vec{j} \times \vec{k}=\vec{l}=-\vec{k} \times \vec{j} ; \vec{k} \times \vec{l}=\vec{j}=-\vec{l} \times \vec{k}\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{j}+a_{3} \vec{k}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{j}+b_{3} \vec{k}\) then \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\vec{l} & \vec{j} & \vec{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{array}\right|\)

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \quad \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right system of non coplanar vectors then \(\)

→ If \(\vec{a}, \vec{b}\) are two vectors then \((\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2}\) = a²b².

Vector Area:
If A is the area of the region bounded by a plane curve and \(\vec{n}\) is the unit vector perpendicular to the plane of the curve such that the direction of curve drawn can be considered anti clock wise then A\(\vec{n}\) is called vector area of the plane region bounded by the curve.

• The vector area of triangle ABC is \(\frac{1}{2} \overline{A B} \times \overrightarrow{A C}=\frac{1}{2} \overrightarrow{B C} \times \overrightarrow{B A}=\frac{1}{2} \overline{C A} \times \overrightarrow{C B}\)
• If \(\vec{a}, \vec{b}, \vec{c}\)are the position vectors of the vertices of a triangle then the vector area of the triangle is \(\frac{1}{2}(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a})\)
• If ABCD is a parallelogram and \(\overrightarrow{A B}=\vec{a}, \quad \overrightarrow{B C}=\vec{b}\) then the vector area of ABCD is \(\vec{a} \times \vec{b}\).
• If ABCD is a parallelogram and \(\overrightarrow{A C}=\vec{a}, \overrightarrow{B C}=\vec{b}\) then vector area of parallelogram ABCD is \(\frac{1}{2}(\vec{a} \times \vec{b})\)
• The vector equation of a line passing through the point A with position vector a and perpendicular to the vectors \(\vec{b} \times \vec{c}\) is \(\vec{r}=\vec{a}+t(\vec{b} \times \vec{c})\).

Scalar Triple Product:

• If \(\vec{a}, \vec{b}, \vec{c}\) are the three vectors, then the real numbers \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is called scalar triple product denoted by \([\vec{a} \vec{b} \vec{c}]\). This is read as ‘box’ \(\vec{a}, \vec{b}, \vec{c}\)
• If V is the volume of the parallelepiped with coterminous edges a, b, c then V = |\([\vec{a} \vec{b} \vec{c}]\)|
• If \(\vec{a}, \vec{b}, \vec{c}\) form the right handed system of vectors then V = \([\vec{a} \vec{b} \vec{c}]\)
• If \(\vec{a}, \vec{b}, \vec{c}\) form left handed system of vectors then -V = \([\vec{a} \vec{b} \vec{c}]\)

Note:

• The scalar triple product is independent of the position of dot and cross. i.e. \(\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{b} \times \vec{c}\)
• The value of the scalar triple product is unaltered so long as the cyclic order remains unchanged
  \([\vec{a} \vec{b} \vec{c}]=[\vec{b} \vec{c} \vec{a}]=[\vec{c} \vec{a} \vec{b}]\)
• The value of a scalar triple product is zero if two of its vectors are equal
  \([\vec{a} \vec{a} \vec{b}]\)= 0 \([\vec{b} \vec{b} \vec{c}]\) = 0
• If a, b, c are coplanar then \([\vec{a} \vec{b} \vec{c}]\) = 0
• If a,b,c form right handed system then \([\vec{a} \vec{b} \vec{c}]\) > 0
• If a,b,c form left handed system then \([\vec{a} \vec{b} \vec{c}]\) < 0
• The value of the triple product changes its sign when two vectors are interchanged
  \([\vec{a} \vec{b} \vec{c}]\) = –\([\vec{a} \vec{c} \vec{b}]\)
• If l,m, n are three scalars \(\vec{a}, \vec{b}, \vec{c}\) are three vectors then \(\)

→ Three non zero non parallel vectors abc nare coplanar iff \([l \vec{a} m \vec{b} \quad n \vec{c}]={lmn}\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)= 0

→ If \(\vec{a}=a_{1} \vec{l}+a_{2} \vec{m}+a_{3} \vec{n}, \vec{b}=b_{1} \vec{l}+b_{2} \vec{m}+b_{3} \vec{n}, \vec{c}=c_{1} \vec{l}+c_{2} \vec{m}+c_{3} \vec{n}\) where \(\vec{l}, \vec{m}, \vec{n}\) form a right handed system of non coplanar vectors, then \([\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
\vec{m} \times \vec{n} & \vec{n} \times \vec{l} & \vec{l} \times \vec{m} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\)

→ The vectors equation of plane passing through the points A, B with position vectors \(\vec{a}, \vec{b}\) and parallel to the vector \(\vec{c}\) is \([\vec{r}-\vec{a} \vec{b}-\vec{a} \vec{c}]\) (or) \([\vec{r} \vec{b} \vec{c}]+[\vec{r} \vec{c} \vec{a}]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)

→ The vector equation of the plane passing through the point A with position vector \(\vec{a}\) and parallel to \(\vec{b}, \vec{c}\) is \([\vec{r}-\vec{a} \vec{b} \vec{c}]\) = 0 i.e. \(\left[\begin{array}{lll}
\vec{r} & \vec{b} & \vec{c}
\end{array}\right]=\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]\)

Skew lines:
Two lines are said to be skew lines if there exist no plane passing through them i.e. the lines lie on two difference planes
Def:- l₁ and l₂ are two skew lines. If P is a point on l₁ and Q is a point on l₂ such that \(\overleftarrow{P Q}\) ⊥ l₁ and PQ ⊥ l₂ then \(\overleftarrow{P Q}\) is called shortest distance and \(\overleftarrow{P Q}\) is called shortest distance line between the lines l₁ and l₂.

The shortest distance between the skew lines \(\vec{r}=\vec{a}+t \vec{b}\) and \(\vec{r}=\vec{c}+t \vec{d}\) is \(\frac{|[\vec{a}-\vec{c} \vec{b} \vec{d}]|}{|\vec{b} \times \vec{d}|}\)

Vector Triple Product:
Cross Product of Three vectors : For any three vectors \(\bar{a}, \bar{b}\) or \(\bar{c}\) then cross product or vector product of these vectors are given as \(\bar{a} \times(\bar{b} \times \bar{c}),(\bar{a} \times \bar{b}) \times \bar{c}\) or \((\bar{b} \times \bar{c}) \times \bar{a}\) etc.

vi. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero vectors and \(\) then b and c are parallel (or collinear) vectors.

vii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are non zero and non parallel vectors then \(\bar{a} \times(\bar{b} \times \bar{c}), \quad \bar{b} \times(\bar{c} \times\bar{a})\) and \(\bar{c} \times(\bar{a} \times \bar{b})\) are non collinear vectors.

viii. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})=\overline{\mathrm{O}}\)

ix. If \(\bar{a}, \bar{b}\) or \(\bar{c}\) are any three vectors then \(\bar{a}(\bar{b} \times \bar{c})+\bar{b} \times(\bar{c} \times \bar{a})+\bar{c} \times(\bar{a} \times \bar{b})\) are coplanar. [since sum of these vectors is zero]

x. \(\bar{a}(\bar{b} \times \bar{c})\) is vector lies in the plane of \(\bar{b}\) and \(\bar{c}\) or parallel to the plane of \(\bar{b}\) and \(\bar{c}\).

Product of Four Vectors:
Dot product of four vectors : The dot product of four vectors a̅, b̅, c̅ and d̅ is given as \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})=(\bar{a} \cdot \bar{c})(\bar{b} \cdot \bar{d})-(\bar{a} \cdot \bar{d})(\bar{b} \cdot \bar{c})=\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{a} \cdot \bar{d} \\
\bar{b} \cdot \bar{c} & \bar{b} \cdot \bar{d}
\end{array}\right|\)

→ Cross product of four vectors : If a̅, b̅, c̅ and d̅ are any four vectors then

→ The vectorial equation of the plane passing through the point a and parallel to the vectors b̅, c̅is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅ and parallel to the vector c̅ is \([\overline{\mathrm{rb}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ The vectorial equation of the plane passing through the points a̅, b̅, c̅ is \([\overline{\mathrm{r}} \overline{\mathrm{b}} \overline{\mathrm{c}}]+[\overline{\mathrm{r}} \overline{\mathrm{c}} \overline{\mathrm{a}}]+[\overline{\mathrm{ra}} \overline{\mathrm{b}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ If the points with the position vectorsa̅, b̅, c̅, d̅ are coplanar, then the condition is \([\overline{\mathrm{a}} \overline{\mathrm{bd}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{d}}]+[\overline{\mathrm{c}} \overline{\mathrm{a}} \overline{\mathrm{d}}]=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]\)

→ Length of the perpendicular from the origin to the plane passing through the points a̅, b̅, c̅ is \(\frac{|[\overline{\mathrm{a} b} \overline{\mathrm{c}}]|}{|\overline{\mathrm{b}} \times \overline{\mathrm{c}}+\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}\)

→ Length of the perpendicular from the point c̅ on to the line joining the points a̅, b̅ is \(\frac{\mid(\overline{\mathrm{a}}-\overline{\mathrm{c}}) \times(\overline{\mathrm{c}}-\overline{\mathrm{b}})}{|\overline{\mathrm{a}}-\overline{\mathrm{b}}|}\)

→ P, Q, R are non collinear points. Then distance of P to the plane OQR is OP. \(\left|\frac{\overline{\mathrm{OP}} \cdot(\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}})}{|\overline{\mathrm{OQ}} \times \overline{\mathrm{OR}}|}\right|\)

→ Perpendicular distance from P(α̅ ) to the plane passing through A(a̅) and parallel to the vectors b and c is

→ Length of the perpendicular from the point c̅ to the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{tb}\) is \(\frac{|(\overline{\mathrm{c}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}|}{|\overline{\mathrm{b}}|}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 4 Addition of Vectors to solve questions creatively.

Intermediate 1st Year Maths 1A Addition of Vectors Formulas

Scalar :
A physical quantity that has only magnitude is called a scalar quantity. All the real numbers will be taken as scalars.

Vector :
A physical quantity that has both magnitude and direction.
e.g. : Velocity acceleration, force, momentum.

Position vector :
Let ‘O’ and ‘P be any points in space. Then OP is called the position vector of the point ‘P w.r.t. origin ‘O’.
Note : \(\overline{A B}\) = Position vector of B- position vector of A’
= \(\overline{O B}-\overline{O A}\)

Coinitial vector:
Vectors having the same initial point are called coinitial vectors.
e.g. : \(\overline{O A}, \overline{O B}, \overline{O C}\) etc.

Unit vector:
A vector whose magnitude is one-unit is called unit vector.
Unit vector in the direction of a is denoted by â = \(\frac{\bar{a}}{|a|}\).
For any non-zero vector \(\bar{a}=|\bar{a}|\) â.

Like vectors :
If two vectors are parallel and having the same direction then they are called like vectors.

Unlike vectors :
If two vectors are parallel and having Opposite direction then they are called unlike vectors.
The position vector of any point C on \(\overline{A B}\) can be taken as λ\(\bar{a}\) + µ\(\bar{a}\), where λ + µ = 1

Angle between vectors:
Let \(\overline{O A}=\bar{a}\) = a, \(\overline{O B}=\bar{a}\) = b be any two non – zero vectors, then angle AOB is defined as angle between vectors \(\bar{a}, \bar{b}\) and is denoted by \((\bar{a}, \bar{b})\) where 0 ≤ \((\bar{a}, \bar{b})\) ≤ 180°.

Addition of vectors (or) Parallelogram Law:
If \(\bar{a}, \bar{b}\) are the adjacent sides of a parallelogram the diagonals which is coinitial with \(\bar{a}, \bar{b}\) is given by \(\bar{a}+ \bar{b}\) and its magnitude is given by
\(|\bar{a}+\bar{b}|=\sqrt{|\bar{a}|^{2}+|\bar{b}|^{2}+2|\bar{a}||\bar{b}| \cos (\bar{a}, \bar{b})}\)

Triangle law:
If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Note : Addition of vectors is of 2 types :

1. Commutative
2. Associative.

(ie) (i) \(\bar{a}+\bar{b}=\bar{b}+\bar{a}\) and (ii) \((\bar{a}+\bar{b})+\bar{c}=\bar{a}+(\bar{b}+\bar{c})\)
Rule: \(|\bar{a}| \sim|\bar{b}| \leq|\bar{a}-\bar{b}| \leq|\bar{a}+\bar{b}| \leq|\bar{a}|+|\bar{b}|\)

Parallel (or) Collinear vector:

1. Two vectors a, b are parallel or collinear, then a = λ. b, where ‘λ’ is a scalar.
2. If \(\bar{a}\) = (a₁, a₂, a₃), b = (b₁, b₂, b₃) are parallel or collinear, then
   \(\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}\)
   Note : Zero vector is parallel to any vector.
3. If three points with position vectors a, b, c are to be collinear. The necessary and sufficient condition is that there exists scalars \(\bar{a}, \bar{b}, \bar{c}\) not all zero, such that
   l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) = 0, l + m + n = 0.
4. If \(\bar{a}, \bar{b}, \bar{c}\) are non-zero, non-collinear vectors such that l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) = 0, then l = 0, m = 0, n = 0.

Linear combination of vectors :
A linear combination of the system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \ldots, \bar{a}_{n}\) is a vector.
r = x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\)
where x₁ x₂, x₃, ………………., x_n are scalars.

Coplanar vectors:
If three or more vectors lie in the same plane (or) parallel to the same plane then they are called coplanar vectors. If one vector can be expressed as a linear combination of the remaining vectors, then the vectors are coplanar vectors.
If \(\bar{a}\) = x\(\bar{b}\) + y\(\bar{c}\), where x, y, are scalars, then \(\bar{a}, \bar{b}, \bar{c}\) are coplanar.

Linearly dependent system of vectors :
A system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \ldots \ldots, \bar{a}_{n}\) is said to
be linearly dependent, if there exists a system of scalars x₁, x₂, x₃, ……………….., x_n not all zero
such that
x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\) = \(\bar{0}\)

• The null vector is linearly dependent.
• Two collinear vectors are linearly dependent.
  \(\bar{a}\) = λ\(\bar{b}\) ⇒ (1)\(\bar{a}\) + (-λ)\(\bar{b}\) = 0
• Any three coplanar vectors are linearly dependent.
  \(\bar{a}\) = x\(\bar{b}\) + y\(\bar{c}\) ⇒ (1)\(\bar{a}\) + (-x)\(\bar{b}\) + (-y)\(\bar{c}\) = 0
• Any four vectors in space from a linearly dependent set of vectors.

Linearly independent system of vectors:
A system of vectors \(\bar{a}_{1}, \bar{a}_{2}, \bar{a}_{3}, \ldots \ldots, \bar{a}_{n}\) is said to be linearly independent, if x₁\(\bar{a}_{1}\) + x₂\(\bar{a}_{2}\) + x₃\(\bar{a}_{3}\) + ………………. + x_n\(\bar{a}_{n}\) = \(\bar{0}\) implies

• Two non-zero, non-collinear vectors are linearly independent.
• Any three non-coplanar vectors are linearly independent. If a, b, c are three non-coplanar vectors and \(\bar{r}\) be any other vector. Then there exists unique scalars x, y, z such that
  \(\bar{r}\) = x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\).

Right handed system of vectors:
Three non-coplanar vectors \(\bar{a}, \bar{b}, \bar{c}\) are said to form a – right-handed system, if the rotation is form \(\bar{a}\) to \(\bar{b}\) in anti-clockwise direction, through an angle less than 180° as seen from the terminal point of c.
If \(\bar{a}, \bar{b}, \bar{c}\) from RHS and \(\bar{b}, \overline{,}, \bar{a}\) and \(\bar{c}, \bar{a}, \bar{b}\) also from RHS.

Left-handed system of vectors:
Three non-coplanar vectors \(\bar{a}, \bar{b}, \bar{c}\) are said to form a left handed system. If the rotation is from \(\bar{a}\) to \(\bar{b}\) in clockwise direction through an angle less than 180° as seen from terminal point of c.

If \(\bar{a}, \bar{b}, \bar{c}\) are in RHS, then
\(\bar{b}, \overline{,}, \bar{a}\); \(\bar{c}, \bar{a}, \bar{b}\) also from R.H.S.
and \(-\bar{a}, \bar{b}, \bar{c} ; \bar{a},-\bar{b}, \bar{c} ; \bar{a}, \bar{b},-\bar{c}\) form L.H.S.

Direction cosines of a vector and direction ratios of a vector:
If α, β, γ are the angles made by a line with +ve directions of x, y, z axes respectively, then cos α, cos β, cos γ are known as the direction cosines of that line.
Generally the direction cosines are denoted by l, m, n and l² + m² + n² = 1.
Any numbers proportional to the direction cosines are known as direction ratios.

→ Unit vector in the direction of \(\bar{a}\) = \(\frac{\bar{a}}{|\bar{a}|}\)

→ Unit vector parallel to the resultant of the vectors \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\) is ± \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\)

→ The vector parallel to the resultant of the vectors \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\) and having magnitude λ is ±λ \(\frac{\bar{a}+\bar{b}+\bar{c}}{|\bar{a}+\bar{b}+\bar{c}|}\)

→ The position vector of the point which divides the line segment joining the points A and B whose position vectors are \(\bar{a}, \bar{b}\) respectively, internally in the ratio, l:m is \(\frac{m \overline{\mathrm{a}}+l \bar{b}}{l+m}\) externally in the ratio l: m is \(\frac{m \bar{a}-l \bar{b}}{m-l}\), m ≠ l.

→ The ratio in which the line joining the points A(x₁ y₁, z₁) and B(x₂, y₂, z₂) is divided by

• xy – plane is – z₁: z₂
• yz – plane is -x₁ : x₂
• zx – plane is -y₁: y₂

→ If ‘c’ is the mid-point of line AB, the position vector of ‘c’ is \(\overline{O C}=\frac{\overline{O A}+\overline{O B}}{2}\)
= \(\frac{\bar{a}+\bar{b}}{2}\)

→ If \(\bar{a}, \bar{b}\) are two unit vectors, then the unit vector along the bisector of the angle between \(\bar{a}, \bar{b}\) is given by
\(\bar{c}=\frac{\bar{a}+\bar{b}}{|\bar{a}+\bar{b}|}\)
or
\(\bar{c}=\frac{\bar{a}-\bar{b}}{|\bar{a}-\bar{b}|}\)

→ The Vector \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) are coplanar or linearly dependent iff \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

→ The Vector \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) are non-coplanar or linearly independent iff \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) ≠ 0

→ The necessary and sufficient condition for four points with position vectors \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\) are coplanar is that there exists scalars l, m, n, p not all zero such that
l\(\bar{a}\) + m\(\bar{b}\) + n\(\bar{c}\) + p\(\bar{da}\) = \(\bar{0}\), l + m + n + p = 0.

→ If \(\bar{a}, \bar{b}, \bar{c}\) are three non-zero, non – coplanar vectors and x, y, z are three scalars such that x\(\bar{a}\) + y\(\bar{b}\) + z\(\bar{c}\) = 0, then x = 0, y = 0, z = 0.
Note : Collinearity implies coplanarity but coplanarity does not imply collinearity.

→ If \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) and if \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) > 0 then \(\bar{a}, \bar{b}, \bar{c}\) are in R.H.S

→ If \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃) and if \(\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) < 0 then \(\bar{a}, \bar{b}, \bar{c}\) are in L.H.S

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\), then |\(\bar{r}\)| = r = \(\sqrt{x^{2}+y^{2}+z^{2}}\) .

→ If l, m, n are the direction cosines of a line, then l² + m² + n² = 1.

→ If a vector makes angles α, β, γ with co-ordinate axes, then

• cos² α + cos² β + cos² γ = 1
• sin² α + sin² β + sin² γ =2

→ The direction ratios of a vector \(\bar{r}\) = a\(\bar{i}\) + b\(\bar{j}\) + c\(\bar{k}\) are a, b, c.

→ If A = (x₁, y₁ z₁), B = (x₂, y₂, z₂) then the direction ratios of a vector
AB are x₂ – x₁, y₂ – y₁, z₂ – z₁

→ If a, b, c are the direction ratios of a vector then direction cosines of a vector are
\(\pm \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \pm \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \pm \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

→ If a, b, c are the direction ratios of a line then λa, λb, λc also become the direction ratios of that line where ‘λ’ is a non-zero scalar.

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\), then the direction cosines of a vector are \(\frac{a}{|\vec{r}|}, \frac{b}{|\vec{r}|}, \frac{c}{|\vec{r}|}\)

→ If \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\) makes angles α, γ with co-ordinate axes respectively, then

cos α = \(\frac{a}{|\vec{r}|}\),
cos β = \(\frac{b}{|\vec{r}|}\)
cos γ = \(\frac{c}{|\vec{r}|}\).

• The direction cosines of x- axis are 1, 0, 0.
• The direction cosines of y – axis are 0, 1,0.
• The direction cosines of z- axis are 0, 0, 1.

→ If a vector make equal angles with co-ordinate axes then direction cosines of a vector are \(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\)

→ If l, m, n are direction cosines of a vector OP and ‘O’ is the origin OP = r, then P = (lr, mr, nr).

→ If 1, m, n are direction cosines of a vector, then the maximum value of lmn is \(\frac{1}{3 \sqrt{3}}\)

→ The maximum value of l + m + n = \(\frac{1}{3 \sqrt{3}}\)

→ The vector equation of a straight-line passing through the point whose position vector is a and parallel to the vector \(\bar{b}\) is \(\bar{r}=\bar{a}+t \bar{b}\), where ‘t’ is parameter.

→ In cartesian form its equation is \(\frac{x-a_{1}}{b_{1}}=\frac{y-a_{2}}{b_{2}}=\frac{z-a_{3}}{b_{3}}\)
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃).

→ The vector equation of a straight-line passing through the point whose position vectors are
a and b is \(\bar{r}\) = (1 – t)\(\bar{a}\) (or) \(\bar{r}=\bar{a}+t(\bar{b}-\bar{a})\); where’t’ is a parameter.

→ In cartesian form \(\frac{x-a_{1}}{b_{1}-a_{1}}=\frac{y-a_{2}}{b_{2}-a_{2}}=\frac{z-a_{3}}{b_{3}-a_{3}}\)
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃).

→ Vector equation of the plane through the points whose position vectors are \(\bar{a}, \bar{b}, \bar{c}\) is \(\bar{r}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}\), where s and t are parameters (or)
\(\bar{r}=\bar{a}+s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})\).
where \(\bar{a}\) = (a₁, a₂, a₃), \(\bar{b}\) = (b₁, b₂, b₃), \(\bar{c}\) = (c₁, c₂, c₃)

→ The vector equation of the plane through the point whose position vector is \(\bar{a}\) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is \(\bar{r}=\bar{a}+s \bar{b}+t \bar{c}\), where ‘s’ and ‘t’ are parameter.
In cartesian form it is \(\left|\begin{array}{ccc}
x-a_{1} & y-a_{2} & z-a_{3} \\
b_{1} & b_{2} & b_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

→ Vector equation of the plane through the point’s whose position vectors are a, b and parallel to vector c is
\(\bar{r}=(1-s) \bar{a}+s \bar{b}+t \bar{c}\)
or
\(\bar{r}=\bar{a}+s(\bar{b}-\bar{a})+t \bar{c}\)
In cartesian form it is \(\left|\begin{array}{ccc}
x-a_{1} & y-a_{2} & z-a_{3} \\
b_{1}-a_{1} & b_{2}-a_{2} & b_{3}-a_{3} \\
c_{1} & c_{2} & c_{3}
\end{array}\right|\) = 0

Scalar:
A quantity which has only magnitude but no directions is called scalar quantity.
Ex: Length, mass, time ………….

Vector:
A quantity which has both magnitude and direction is called a vector quantity
Ex: Displacement, Velocity, Force ………………
A vector can also be denoted by a single letter \(\vec{a}, \vec{b}, \vec{c}\) …….. or bold letter a, b, c
Length of \(\vec{a}\) is dinded by |\(\vec{a}\)|. Length of \(\vec{a}\) is called magnitude of \(\vec{a}\).

Zero vector (Null Vector):
The vector of length O and having any direction is called null vector. It is denoted by \(\bar{O}\)

Note:

• If A is any point in the space then \(\overrightarrow{A A}=\vec{O}\)
• A non zero vector is called a proper vector.

Free vector:
A vector which is independent of its position is called free vector

Localised vector:
If \(\vec{a}\) is a vector P is a point then the ordere of pair, (P,a) is called localized vector at P

Multiplication of a vector by a scalar:

• Let m be any scalar and \(\vec{a}\) be any vector then the vector m \(\vec{a}\) is defined as
• Length of m \(\vec{a}\) is |m| times of length of \(\vec{a}\) i.e. |m\(\vec{a}\)| = |m||a|
• The line of support of m\(\vec{a}\) is same or parallel to that of \(\vec{a}\)

The sense the direction of m\(\vec{a}\) is same as that of \(\vec{a}\) if m is positive, the direction of m\(\vec{a}\) is opposite to that \(\vec{a}\) if m is negative

Note:

• o\(\vec{a}\) = o
• m\(\vec{0}\) = \(\vec{0}\)
• (mn)\(\vec{a}\) = m(n\(\vec{a}\)) = n(m\(\vec{a}\))
• (-1)\(\vec{a}\)

Negative of a vector:
\(\vec{a}\) ,\(\vec{b}\) are two vectors having same length but their directions are opposite to each other then each vector is called the negative of the other vector.
Here \(\vec{a}\) = –\(\vec{b}\) and \(\vec{b}\) = –\(\vec{a}\)

Unit vector:
A vector whose magnitude is unity is called unit vector.
Note: If \(\vec{a}\) is any vector then unit vector is \(\frac{\vec{a}}{|\vec{a}|}\) this is denoted by â {read as ‘a’ cap}

Collinear or parallel vectors:
Two or more vectors are said to the collinear vectors if the have same line of support. The vectors are said to be parallel if they have parallel lines of support.

Like parallel vectors:
Vectors having same direction are called like parallel vectors.

Unlike parallel vectors:
Vectors having different direction are called unlike parallel vectors.

Note:

• If \(\vec{a}\), \(\vec{b}\) are two non-zero collinear or parallel vectors then there exists a non zero scalar m such that \(\vec{a}\) = m\(\vec{b}\)
• Conversly if there exists a relation of the type \(\vec{a}\) = m\(\vec{b}\) between two non zero two non zero vectors a, b then a, b must be parallel or collinear.

Cointialvectors:
The vectors having the same initial point are called co-initial vectors.

Co planar vectors:
Three or more vectors are said to be coplanar if they lie on a plane parallel to same plane. Other wise the vectors are non coplanar vectors.

Angle between two non-zero vectors:
Let \(\overline{O A}=\vec{a}\) and \(\overline{O B}=\vec{b}\) be two non-zero vectors. Then the angle between \(\vec{a}\) and \(\vec{b}\) is defined as that angle AOB where O ≤ ∠AOB ≤ 180°

The angle between \(\vec{a}\) and \(\vec{b}\) is denoted by (\(\vec{a}\), \(\vec{b}\))

Note:

• If \(\vec{a}\) and \(\vec{b}\)are any two vectors such that (\(\vec{a}\), \(\vec{b}\)) = 0 then o ≤ 0 ≤ 180° {this is the range of θ i.e. angle beween vectors}
• \((\vec{a}, \vec{b})=(\vec{b}, \vec{a})\)
• \((\vec{a},-\vec{b})=(-\vec{a}, \vec{b})\) = 180° – \((\vec{a}, \vec{b})\)
• \((-\vec{a},-\vec{b})=(\vec{a}, \vec{b})\)

• If k > 0; 1 > 0 then \((k \vec{a}, l \vec{b})=(\vec{a}, \vec{b})\)
• If \((\vec{a}, \vec{b})\) = 0 then \(\vec{a}, \vec{b}\) are like parallel vectors
• If \((\vec{a}, \vec{b})\) = 180° then \(\vec{a}, \vec{b}\) are unlike parallel vectors
• If \((\vec{a}, \vec{b})\) = 90° then \(\vec{a}, \vec{b}\) are orthogonal or perpendicular vectors.

Addition of vectors:
Let \(\vec{a}\) and \(\vec{b}\) be any two given vectors. If three points O, A, B are taken such that \(\overline{O A}=\vec{a}\), \(\overrightarrow{A B}=\vec{b}\) then vectors \(\) is called the vector sum or resultant of the given vectors a and b we write \(\overline{O B}=\overline{O A}+\overline{A B}=\vec{a}+\vec{b}\)

Triangular law of vectors:
Trianglular law states that if two vectors are represented in magnitude and direction by two sides of a triangle taken in order, then their sum or resultant is represented in magnitude and direction by the third side of the triangle taken in opposite direction.

Position vector:
Let O be a fixed point in the space called origin. If P is any point in the space then \(\) is called position vector of P relative to O.

Note:
If \(\vec{a}\) and \(\vec{b}\) be two non collinear vectors, then there exists a unique plane through a,b this plane is called plane generated by a, b. If \(\overline{O A}=\vec{a} ; \overline{O B}=\vec{b}\) then the plane generated by \(\vec{a}\), \(\vec{b}\) is denoted by \(\overline{A O B}\).

• Two non zero vectors a, b are collinear if m\(\vec{a}\) + n\(\vec{b}\) = 0 for some scalars m,n not both zero
• Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of the points A, B, C respectively. Then A, B, C are collinear iff m\(\vec{a}\) + n\(\vec{b}\) + p\(\vec{c}\) = 0 for some scalars m, n, p not all zero such that m + n + p = 0
• Let \(\vec{a}, \vec{b}\) be two non collinear vectors, If \(\vec{r}\) is any vector in the plane generated by a,b then there exist a unique pair of real numbers x, y such that \(\vec{r}\) = x\(\vec{a}\) + y\(\vec{b}\).
• Let \(\vec{a}, \vec{b}\) be two non collinear vectors. If \(\vec{r}\) is any vector such that \(\vec{r}\) = xa + yb for some real numbers then \(\vec{r}\) lies in the plane generated by a, b.
• Three vectors \(\vec{a}, \vec{b}, \vec{c}\) coplanar iff x\(\vec{a}\) + y\(\vec{b}\) + z\(\vec{c}\) = 0 for some scalars x, y, z not all zero.
• Let \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) be the position vectors of the points A, B, C, D in which no three of them are collinear. Then A, B, C, D are coplanar iff m\(\vec{a}\) + n\(\vec{b}\) + p\(\vec{c}\) + q\(\vec{d}\) = 0 for some scalars m,n,p,q not all zero such that m + n + p + q = 0
• If \(\vec{a}, \vec{b}, \vec{c}\) are three non coplanar vectors and \(\vec{r}\) is any vector then there exist a unique traid of real numbers. x, y, z such that \(\vec{r}\) = x\(\vec{a}\) + y\(\vec{b}\) + z\(\vec{c}\).

Right handed system of orthonormal vectors
A triced of three non-coplanar vectors\(\vec{i}, \vec{j}, \vec{k}\) is said to be a right hand system of orthonormal triad of vectors if

• \(\vec{i}, \vec{j}, \vec{k}\) from a right handed system
• \(\vec{i}, \vec{j}, \vec{k}\) are unit vectors
• (i, j) = 900 = \((\vec{j}, \vec{k})=(\vec{k}, i)\)

Right handed system:
Let \(\overline{O A}=\vec{a}, \overline{O B}=\vec{b}\) and \(\overrightarrow{O C}=\vec{c}\) be three non coplanar vectors. If we observe from the point C that a rotation from OA to OB through an angle not greater than 1800 is in the anti clock wise direction then the vectors a,b, c are said to form ‘Right handed system’.

Left handed system:
If we observe from C that a rotation from OA to OB through an angle not greater than 1800 is in the clock -wise direction then the vectors a,b, c are said to form a “Left handed system”.
If \(\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}\) then \(|\vec{r}|=\sqrt{x_{2}+y_{2}+z_{2}}\).

Direction Cosines:
If a given directed line makes angles α, β, γ, with positive direction of axes of x,y, and z respectively then cos α,cos β,cos γ are called direction cosines of the line and these are denoted by l,m,n.

Direction ratios:
Thre real numbers a,b,c are said to be direction ratios of a line if a:b:c = l:m:n where l,m,n are the direction cosines of the line.

Linear combination:
Let \(\vec{a}_{1}, \bar{a}_{2}, \overline{a_{3}} \ldots \ldots \overline{a_{n}}\) be n vectors and l₁,l₂,l₃….l_n be n scalars then \(l_{1} \bar{a}_{1}+l_{2} \overline{a_{2}}+l_{3} \overline{a_{3}}+\ldots+l_{n} \overline{a_{n}}\) is called a. linear combination of \(\vec{a}_{1}, \bar{a}_{2} \ldots \ldots . \overline{a_{n}}\)

Linear dependent vectors:
The vectors \(\bar{a}_{1}, \overline{a_{2}} \ldots \ldots \overline{a_{n}}\) are said to be linearly dependent if there exist scalars l₁,l₂,l₃….l_n not all zero such that \(_{1} \vec{a}_{1}+l_{2} \overline{a_{2}}+\ldots+l_{n} \overline{a_{n}}]\) = 0

Linear independent The vectors a₁, a₂, a₃ ……………. a_n are said to be linearly independt if l₁, l₂, l₃ ……………. l_n are scalars, \(l_{1} \vec{a}_{1}+l_{2} \overline{a_{2}}+\ldots l_{n} \overline{a_{n}}\) = 0
⇒ l₁ = 0 l₂ = o…. l_n = o

• Let \(\vec{a}, \vec{b}\) be the position vectors of A, B respectively the position vector of the point P which divides \(\overline{A B}\) in the ratio m:n is \(\frac{m \vec{b}+n \vec{a}}{m+n}\). Conversely the point P with position vector \(\frac{m \vec{b}+n \vec{a}}{m+n}\) lies on the lines \(\overline{A B}\) and divides \(\overline{A B}\) in the ratio m:n.
• The medians of a triangle are concurrent. The point of concurrence divides each median in the ratio 2:1
• Let ABC be a triangle and D be a point which is not in the plane of \(\overline{A B C}\) the lines joining O, A, B,C with the centroids of triangle ABC, triangle BCD, triangle CDA and triangle DAB respectively are concurrent and the point of concurrence divides each line segment in the ratio 3:1
• The equation of the line passing through the point A = (x₁, y₁, z₁) and parallel to the vector b = (l, m, n) is \(\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}\) = t
• The equation of the line having through the points A(x₁, y₁, z₁), B(x₂, y₂, z₂) is \(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\) = t
• The unit vector bisecting the angle between the vectors \(\vec{a}, \vec{b}\) is \(\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\)
• The internal bisector the angle between \(\vec{a}, \vec{b}\) is \(\frac{\overline{O P}}{O P}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}\) are concurrent

Theorem 1:
The vector equation of a line parallel to the vector \(\vec{b}\) and passing through the point A with position vector \(\vec{a}\) is \(\vec{r}=\vec{a}+t \vec{b}\) where t is a scalar.
Proof:
Let \(\overline{O A}=t \vec{a}\) be the given point and \(\overline{O P}=\vec{r}\) be any point on the line
\(\overline{A P}=t \vec{b}\) where ‘t’ is a scalar
\(\vec{r}-\vec{a}=t \vec{b} \Rightarrow \vec{r}=\vec{a}+t \vec{b}\)

Theorem 2:
The vector equation of the line passing through the points A, B whose position vectors \(\vec{a}, \vec{b}\) respectively is \(\vec{r}=\vec{a}(1-t)+t \vec{b}\) where t is a scalar.
Proof:
let P a a point on the line joining of A, B

Theorem 3:
The vector equation of the plane passing through the point A with position vector \(\vec{b}, \vec{c}\) and parallel to the vectors \(\vec{a}\) is \(\vec{r}=\vec{a}+s \vec{b}+t \vec{c}\) where s,t are scalars
Proof:
Given that \(\overline{O A}=\vec{a}\)
Let \(\overline{O P}=\vec{r}\) be the position vector of P
\(\overrightarrow{A P}, \vec{b}, \vec{c}\) are coplanar
∴ \(\overrightarrow{A P}=s \vec{b}+t \vec{c} \Rightarrow \vec{r}-\vec{a}=s \vec{b}+t \vec{c}\)
∴ \(\vec{r}=\vec{a}+s \vec{b}+t \vec{c}\)

Theorem 4:
The vector equation of the plane passing through the points A, B with position vectors \(\vec{a}, \vec{b}\) and parallel to the vector \(\vec{c}\) is \(\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c}\)
Proof:
Let P be a point on the plane and \(\overline{O P}=\vec{r}\)
\(\overline{O A}=\vec{a} \overline{O B}=\vec{b}\) be the given points \(\overline{A P}, \overrightarrow{A B}, \vec{C}\) are coplanar
\(\overrightarrow{A P}=s \overrightarrow{A B}+t \vec{C}\)
\(\vec{r}-\vec{a}=s(\vec{b}-\vec{a})+t \vec{c}\)
\(\vec{r}=(1-s) \vec{a}+s \vec{b}+t \vec{c}\)

Theorem 5:
The vector equation of the plane passing through the points A, B, C having position vectors \(\vec{a}, \vec{b}, \vec{c}\) is \(\vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c}\) where s,t are scalars
Proof:
Let P be a point on the plane and \(\overline{O P}=\vec{r}\)
\(\overline{O A}=\vec{a} \overline{O B}=\vec{b}, \overline{O C}=\bar{c}\)
Now the vectors \(\overline{A B}, \overline{A C}, \overline{A P}\) are coplanar.
Therefore, \(\overline{A P}=s \overline{A B}+t \overline{A C}\), t,s are scalars.
\(\bar{r}-\bar{a}=s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})\)
\(\vec{r}=(1-s-t) \vec{a}+s \vec{b}+t \vec{c}\)

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 3 Matrices to solve questions creatively.

Intermediate 1st Year Maths 1A Matrices Formulas

→ An ordered rectangular array of elements is called a matrix.

→ A matrix in which the number of rows is equal to the number of columns, is called a square matrix. Otherwise, it is called a rectangular matrix. In a square matrix, an element a_ij is in principal diagonal, if i = j

→ If each non-diagonal element of a square matrix is equal to zero, then it is called a diagonal matrix.

→ If each non-diagonal element of a square matrix is equal to zero and each diagonal element is equal to a scalar, then it is called a scalar matrix.

→ If each non-diagonal element of a square matrix is equal to zero and each diagonal element is equal to 1, then that matrix is called a unit matrix or Identity matrix.

→ If A is a square matrix then the sum of elements in the principal diagonal of A is called trace of A.

• Matrix addition is commutative.
• Matrix addition is associative.
• Matrix multiplication is associative.
• Matrix multiplication is distributive over matrix addition.

→ A square matrix A is said to be an idempotent matrix if A² = A.

→ A square matrix A is said to be an involuntary matrix if A² = I.

→ A square matrix A is said to be a nilpotent matrix, if there exist a positive integer n such that Aⁿ = 0. If n is the least positive integer such thatAn = 0, then n is called index of the nilpotent matrix A.

→ The matrix obtained by interchanging rows and columns is called transpose of the given matrix. Transpose of A is denoted by A^T (or) A’.

→ For any matrices A and B

• (A^T)^T = A
• (A + B)^T = A^T + B^T (A and B are same order)
• (AB)^T = B^TA^T. (If A and B are of orders m xn and n xp respectively)

→ A square matrix A is said to be symmetric if A^T = A.

→ A square matrix A is said to be skew symmetric if A^T = A.

→ Every square matrix can be uniquely expressed as a sum of symmetric matrix and a skew symmetric matrix.

→ The minor of an element in the square matrix of order ‘3’ is defined as the determinant of 2 × 2 matrix, obtained after deleting the row and the column in which the element is present.

→ The cofactor of an element in the i^th row and j^th column of 3 × 3 matrix is defined as its minor multiplied by (-1)^i+j.

→ The sum of the products of the elements of any row or column with their corresponding cofactors is called the determinant of a matrix.

• A square ‘A’ is said to be a singular matrix if det A = 0.
• A square A’ is said to be a non-singular matrix if det A ≠ 0.

→ The transpose of the matrix obtained by replacing.the elements of a square matrix A by the corresponding cofactors is called the adjoint matrix of A and it is denoted by adj (A).

• A square matrix ‘A is said to be an invertible matrix if there exists a square matrix B such that AB = BA = I the matrix B is called inverse of A and it is denoted by A^-1.
• If A is an invertible then A^-1 = 1.

→ A square matrix A is non-singular if A is invertible.

• If A and B are non-singular matrices of same type then Adj (AB) = (Adj B) (Adj A).
• If A is a square matrix of type n then det (Adj A) = (det A)^n-1.

→ A matrix obtained by deleting some rows or columns (or both) of a matrix is called a submatrix.

→ Let Abe a non-zero matrix, the rank of A is defined as the maximum of the orders of the non-singular square submatrices of A. The rank of a null matrix is zero. The rank of a matrix A is denoted as rank (A) or P (A).

→ A system of linear equations is

• Consistent, if it has a solution
• Inconsistent, if it has no solution

→ Non homogeneous system

• a₁x + b₁y + c₁z = d₁
• a₂x + b₂y + c₂z = d₂
• a₃x + b₃y + c₃z = d₃

→ The above system of equations has

• aunique solution if rank (A) = Rank [AD] = 3
• infinity many solutions, if rank (A) = Rank ([AD]) < 3
• no solution, if rank A ≠ Rank ([AD])

→ Homogeneous system of equations

• a₁x + b₁y + c₁z = d₁
• a₂x + b₂y + c₂z = d₂
• a₃x + b₃y + c₃z = d₃

→ The above system has

• Trival solution x = y = z = 0 only if rank (A) = 3
• infinitely many non-trival solutions if rank (A) < 3.

→ A matrix is an arrangement of real or complex numbers into rows and columns so that all the rows (columns) contain equal no. of elements.

→ If a matrix consists of ‘m’ rows and ‘n’ columns, then it is said to be of order m × n.

→ A matrix of order n × n is said to be a square matrix of order n.

→ A matrix (a_ij)_m×n is said to be a null matrix if a_ij = 0 for all i and j.

→ Two matrices of the same order are said to be equal if the corresponding elements in the matrices are all equal.

→ A matrix (a_ij)_n×n is a diagonal matrix a_ij = 0 for all i ≠ j

→ A matrix (a_ij)_n×n is a scalar matrix if a = 0 for all i ≠ j and a_ij = k (constant) for i = j

→ A matrix (a_ij)_n×n is said to be a unit matrix of order n, denoted by I_n if a_ij = 1, when i = j and a_ij = 0 when i ≠ j
Ex: I₂ = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
I₃ = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

→ If A = (a_ij)_m×n, B = (b_ij)_m×n, then A + B = (a_ij + b_ij)_m×n

→ Matrix addition is commutative and associative

→ Matrix multiplication is not commutative but associative

→ If A is a matrix of order m × n, then AI_n = I_mA = A(AI = IA = A)

→ If AB = CA = I, then B = C

→ If A = (a_ij)_m×n, then A ^T = (a_ij)_n×m

→ (KA)^T = KA^T, (A + B)^T = A^T + B^T, (AB)^T = B^T.A^T

→ A(B + C) = AB + AC, (A + B)C = AC + BC

→ A square matrix is said to be “non-singular” if detA ≠ 0

→ A square matrix is said to be “singular” if detA = 0

→ If AB = 0, where A and B are non-zero square matrices, then both A are singular.

→ A minor of any element in a square matrix is determinant of the matrix obtained by omitting the row and column in which the element is present.

→ In (a_ij)_n×n, the cofactor of a_ij is (-1)^i+j × (minor of a_ij).

→ In a square matrix, the sum of the products of the elements of any row (column) and the corresponding cofactors is equal to the determinant of the matrix.

→ In a square matrix, the sum of the products of the elements of any row (column) and the corresponding cofactors of any other row (column) is alway s zero.

→ If A is any square matrix, then A adjA = adjA. A = detA. I

→ If A is any square matrix and there exists a matrix B such that AB = BA = I, then B is called the inverse of A and denoted by A^-1.

→ AA^-1 = A^-1A = I.

→ If A is non-singular, then A^-1 = \(\frac{{adj} A}{{det} A}\) (or) adj A = |A|AA^-1

→ If A = \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\), then A^-1 = \(\frac{1}{a d-b c}\left(\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right)\)

→ (A^-1)^-1 = A, (AB)^-1 = B^-1.A^-1, (A^-1)^T =( A^T)^-1; (ABC….)^-1 = C^-1B^-1A^-1.

Theorem:
Matrix multiplication is associative. i.e. if conformability is assured for the matrices A, B and C, then (AB)C = A(BC).
Proof:

Theorem:
Matrix multiplication is distributive over matrix addition i.e. if conformability is assured for the matrices A, B and C, then
(i) A (B + C) = AB + AC
(ii) (B + C) A = BA + CA
Proof:
Let A = (a_ij)_m×n, B = (b_jk)_n×p C = (c_ki)_n×p
B + C = (d_jk)_n×p, where d_jk = b_jk + c_jk

∴ A(B + C) = AB + AC
Similarly we can prove that
(B + C) = BA + CA.

Theorem:
If A is any matrix, then (A^T)^T = A.
Proof:
Let A = (a_ij)_m×n
A^T = (a’_jk)_n×m, where a’_ji = a_ij
(A^T)^T = (a”_ji)_m×n, where a”_ij = a_ji
a”_ij = a’_ji = a_ij
∴ (A^T)^T = A

Theorem:
If A and B are two matrices o same type, then (A + B)^T = A^T + B^T.
Proof:
Let A = (a_ij)_m×n, B = (b_ij)
A + B = (c_ij)_m×n,where c_ij = a_ij + b_ij
(A + B)^T = (c’_ji)_n×m. c’_ji = c_ij
A^T = (a’_ji)_n×m,where a’_ji = a_ij
B^T = (b_ji)_n×m. where. b’_kj = b_jk
A^T + B^T = (d_ji)_n×m, where d_ji = a’_ji + b’_ji
c’_ji = c_ij = a_ij + b_ij = a’_ji + b’_ji = d_ji
∴(A + B)^T = A^T + B^T

Theorem:
If A and B are two matrices for which conformability for multiplication is assured, then (AB)^T = B^TA^T.
Pr0of:
Let A = (a_ij)_m×n, B = (b_ji)_n×p
AB = (c_ik)_m×p, where c_ik = \(\sum_{j=1}^{n}\) a_ijb_jk
(AB)^T = (c_ki)_p×m,where c_ki = c_ik
A^T = (a_ji)_n×m,where a_ji = a_ij
B^T = (b_kj)_p×n, where b_kj = b_jk
B^T . A^T = (d_ki )_p×m, where d_ki= \(\sum_{j=1}^{n}\) b_kja_ji
c’ki = cik = \(\sum_{j=1}^{n}\) a_ijb_jk= \(\sum_{j=1}^{n}\) b_kja_jid_ki
∴ (AB)^T = B^TA^T

Theorem:
If A and B are two invertible matrices of same type then AB is also invertible and (AB)^-1 = B^-1A^-1.
Proof:
A is invertible matrix ⇒ A^-1 exists and AA^-1 = A^-1A = I.
B is an invertible matrix ⇒ B^-1 exists and
BB^-1 = B^-1B = I
Now (AB)(B^-1A^-1) = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I
(B^-1A^-1)(AB) = B^-1(A^-1A)B ∴ AB is invertible and
= B^-1IB = B^-1B = I
(AB)(B^-1A^-1) = (B^-1A^-1) = (B^-1A^-1)(AB) = 1
(AB)^-1 = B^-1A^-1.

Theorem:
If A is a non-singular matrix then A is invertible and A^-1 = \(\frac{{Adj} A}{{det} A}\).
Proof:
Let A = \(\left[\begin{array}{lll}
\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\
\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2} \\
\mathrm{a}_{3} & \mathrm{~b}_{3} & \mathrm{c}_{3}
\end{array}\right]\) be a non – singular matrix.
∴ det A ≠ 0

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 2 Mathematical Induction to solve questions creatively.

Intermediate 1st Year Maths 1A Mathematical Induction Formulas

Principle of finite mathematical induction:
Let S be a subset of N such that

• 1 ∈ S
• For any k ∈ N, k ∈S ⇒ (k + 1) ∈ S

Then S = N

Principle of complete mathematical induction:
Let S be a subset of N such that

• 1 ∈ S
• For any k ∈ N {1, 2, 3 … k} ⊆ S
  ⇒ (k + 1) ∈ S

Then S = N

Steps to prove a statement using the principle of mathematical induction :

• Basis of induction : Show that P(1) is true
• Inductive hypothesis : For k > 1, assume that P(k) is true
• Inductive Step : Show that P(k + 1) is true on the basis of the inductive hypothesis.

Principle of finite Mathematical Induction:
Let {P(n) / n ∈ N} be a set of statements. If

• p(1) is true
• p (m) is true ⇒ p (m+1) is true ; then p (n) is true for every n ∈ N.

Principle of complete induction:
Let {P (n) / n N} be a set of statements. If p (1) is true and p(2), p(3) …. p (m-1) are true ⇒ p(m) is true, then p (n) is true for every n e N.

Note:

• The principle of mathematical induction is a method of proof of a statement.
• We often use the finite mathematical induction, hence or otherwise specified the mathematical induction is the finite mathematical induction.

Some important formula:

• Σn = \(\frac{n(n+1)}{2}\)
• Σn² = \(\frac{n(n+1)(2 n+1)}{6}\)
• Σn³ = \(\frac{n^{2}(n+1)^{2}}{4}\)
• a, (a + d), (a + 2d), ……….. are in a.p
  n th term t_n = a + (n – 1)d, sum of n terms S_n = \(\frac{n}{2}\)[ 2a + (n – 1)d] = \(\frac{n}{2}\)[a + l]
  a = first term, l= last term.
• a, ar, ar², ………… is a g.p
  Nth terms t_n = a.r^n-1 a = 1st term, r = common ratio
• Sum of n terms s_n = a\(\frac{\left(r^{n}-1\right)}{r-1}\); r > 1 = a\(\left(\frac{1-r^{n}}{1-r}\right)\); r < 1

Use these Inter 1st Year Maths 1A Formulas PDF Chapter 1 Functions to solve questions creatively.

Intermediate 1st Year Maths 1A Functions Formulas

Function:
Let A and B be non-empty sets and f be a relation from A to B. If for each element of A, there exists a unique element b e B such that (a,b) ∈ f, then f is called a function or mapping from A to B (or A into B). It is denoted by f: A → B. The set A is called the domain of f and B is called the co-domain off.

Illustration: Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16, 25}. Consider the relation f(x) = x², then f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16. Clearly, each element in A has a unique image in B. So, f: A → B.
f = {(1, 1) (2, 4) (3, 9) (4, 16)} is a function from A to. B.
Clearly Domain (f) = {1,2, 3, 4} and Range (f) = {1, 4, 9, 16}
Note : The number of functions that can be defined from, a non-empty finite set A into a non-empty finite set B is [n(B)]^n(A).

If f: A → B a function then f(A) = {f(a) / a ∈ A} is called the range off. It is a subset of B (ie) f(A) ≤ B.

One – one function (Injection):
Let f: A B, then f is said to be one-one function if different elements of A have different f- images in B. Thus f: A → B is one-one. (μ) f: A B is an injection <=> ay a2 e A and f(a7) = f(aj implies that a₁ = a₂

Illustration:
LetA = {1, 2, 3} and B = {2, 4, 6}. Consider f: A → B. f(x) = 2x then f(1) = 2, f(2) = 4 and f(3) = 6. Clearly f is a function from A to B such that different elements in A have different f – images in B.
∴ f: {(1, 2) (2, 4) (3, 6)} is one – one.

Note : The number of one-one functions that can be defined from a non-empty finite set A into a non-empty finite set B is n(B) P_n(A) if n(B) ≥ n(A) and zero if n(B) < n(A).

Onto function (Surjection):
Let f: A → B. If every element of B occurs as the image of at least one element of in A, then fis an onto function. Thus f: A → B is surjection iff for each b e B, 3 a e A such that f(a) = b. Clearly fis onto ⇔ Range (f) = B.

Illustration; Let A = {-1, 7, 2, -2}, B = {1, 4} and Let f: A → B, be a function defined by f(x) = x² then fis onto because f(A) = {f(-1), f(1), f(2), f(-2)} = {1, 4} = B.
Note : The number of onto functions that can be defined from a non – empty finite set A onto a two element set B is 2^n(A) – 2 if n(A) ≥ 2 and zero if n(A) < 2.

Bijective function :
A function f: A → B is a bijection if

• It is one – one i.e., f(a) – f(b) ⇒ a = b ∀ a, b ∈ A.
• It is onto i.e., ∀ b ∈ B ∃ a ∈ A such that f(a) = b.

Note : Number of bijections that can be defined from A to B is [n(A)]!, [n(A) = n(B)].

If f: A → B is a bijection then the relation f^-1 = {(b, a) / (a, b) ∈ f} is a function from B to A and is called the inverse function off.

Constant function :
Let f: A B defined in such a way that all the elements of A have the same f- image in B, then f is said to be a constant function.

Illustration : Let A – {1, 2, 3} and B = {6, 7, 8}. Let f: A → B
f(x) = 6 ∀ x ∈ A i.e., f = {(1, 6) (2, 6) (3, 6)} is a constant function.
The range of a constant function is a singleton set.

Identity function :
Let A be a non-empty set then the function f: A → A defined by, f(x) = x ∀ x ∈ I_A is called the identity function on A and is denoted by I_A. The identity function is bijective.

Let f: A → B, g: B → C be functions. Then gof: A → C is a function and (gof) (a) = g If (a)] ∀ a ∈ A, is called composite of ‘g’ with ‘f.

If f : A → B, g B → C are bijections so is (go f) : A → C and (gof)^-1 = f^-1o g^-1.

• If f: A → B is a bijection, then fof^-1 = I_B and f^-1of = I_A.
• If f: A → B, g: B → C such that go f = I_A fog = I_B then f is a bijection and g = f^-1

Let A be a non-empty subset of R such that – x ∈ A, for all x ∈ A and f: A → R.

• If f(-x) = f(x), ∀ x ∈ A then fis called an EVEN function.
• If f(-x) = – f(x), ∀ x ∈ A then f is called an ODD function.

Functions:
Def 1:
A relation f from a set A into a set B is said to be a function or mapping from A into B if for each x ∈ A there exists a unique y ∈ B such that (x, y) ∈ f. It is denoted
by f : A → B.
Note: Example of a function may be represented diagrammatically. The above example can be written diagrammatically as follows.

Def 2:
A relation f from a set A into a set B is a said to be a function or mapping from a into B if
i) x ∈ A ⇒ f (x) ∈ B
ii) x₁, x₂ ∈ A, x₂ ⇒ f (x₁) = f (x₂)

Def 3:
If f : A → B is a function, then A is called domain, B is called codomain and f (A) = {f (x): x ∈ A} is called range of f.

Def 4:
A function f : A → B if said to be one one function or injection from A into B if different element in A have different f-images in B.
Note:

• A function f : A → B is one one if f(x₁, y) ∈ f,(x₂, y) ∈ f ⇒ x₁ = x₂.
• A function f : A → B is one one iff x₁, x₂ ∈ A, x₁ ≠ x₂ ⇒ f (x₁) ^ f (x₁)
• A function f : A → B is one one iff x₁, x₂ ∈ A, f (x₁) = f (x₂) ⇒ x₁ = x₂
• A function f : A → B which is not one one is called many one function
• If f : A → B is one one and A, B are finite then n(A) < n(B).

Def 5:
A function f : A → B is said to be onto function or surjection from A onto B if f(A) = B.

Note:

• A function f : A → B is onto if y e B U ⇓ ∃x ∈ A ∋ f (x) = y .
• A function f : A → B which is not onto is called an into function.
• If A, B are two finite sets and f : A → B is onto then n(B) ≤ n(A).
• If A, B are two finite sets and n (B) = 2, then the number of onto functions that can be defined from A onto B is 2^{n( A)} – 2.

Def 6:
A function f : A → B is said to be one one onto function or bijection from A onto B if f : A → B is both one one function and onto function.

Theorem: If f : A → B, g : B → C are two functions then the composite relation gof is a function a into C.
Theorem: If f : A → B, g : B → C are two one one onto functions then gof : A → C is also one one be onto.
i) Let x₁, x₂ ∈ A and f (x₁) = f (x₂).
x₁,x₂ ∈ A, f : A → B ⇒ f (x₁), f (x₂) ∈ B
f (x₁), f (x₂) ∈ B → C, f (x₂) ⇒ g[f (x₁)] = g[f (x₂)] ⇒ (gof)(x₁) = (gof)(x₂)
x₁, x₂ ∈ A,(gof)(x₁) = (gof): A → C is one one ⇒ x₁ = x₂
x₁, x₂ ∈ A, f (x₁) = f (x₂) ⇒ x₁ = x₂.
∴ f: A → B Is one one.

ii) Proof: let z ∈ C,g : B → C is onto B y ∈ B ∃:g (y) = z y ∈ Bf : A → B is onto
∃x ∈ A ∋ f (x) = y
G {f(x)} = t
(g o f) x = t
∀ z ∈ CB x ∈ A ∋ (gof)(x) = z.
∴ g is onto.

Def 7:
Two functions f : A → B, g : C → D are said to be equal if

• A = C, B = D
• f (x) = g (x) ∀ x ∈ A. It is denoted by f = g

Theorem:
If f : A → B, g : B → C , h: C → D are three functions, then ho(gof) = (hof )of

Theorem:
if A is set, then the identify relation I on A is one one onto.

Def 8:
If A is a set, then the function I on A defined by I(x) = x ∀ x ∈ A, is called identify function on A. it is denoted by IA.

Theorem: If f : A → B and IA, IB are identify functions on A, B respectively then
foIA = IBof = f .
Proof:
I_A: A → A , f: A → B ⇒ foI_A: A → B
f : A → B , I_B : B → B ⇒ I_Bof: A → B
(foIA)(x) = f {IA(x)} = f (x), ∀x ∈ A ∴ f0I_A = f
(IBof)(x) = IB{f (x)} = f (x), ∀ x ∈ A ∴ I_Bof = f
∴ foI_A = I_Bof = f

Def 9:
If f : A → B is a function then {(y, x) ∈ B × A:(x, y) ∈ f} is called inverse of f. It is denoted by f^-1.

Def 10:
If f : A → B is a bijection, then the function f^-1: B → A defined by f^-1(y) = x iff f (x) = y ∀ y ∈ B is called inverse function of f.

Theorem:
If f : A → B is a bijection, then f^-1 of = IA, fof^-1 = IB
Proof:
Since f : A → B is a bijection f^-1: B → A is also a bijection and
f^-1 (y) = x ⇔ f (x) = y ∀ y ∈ B
f : A → B, f^-1: B → A ⇒ f^-1 of: A → A
Clearly I_A : A → A such that I_A (x) = x, ∀ x ∈ A.

Let x ∈ A
x ∈ A, f : A → B ⇒ f (x) ∈ B
Let y = f(x)
y = f (x) ⇒ f^-1(y) = x
(f -1 of)(x) = f ^-1[ f (x) = f^-1( y) = x = I_A (x)
(f^-1 of) (x) = IA (x) ∀ x ∈ A f^-1of = I_A
f^-1: B → A, f: A → B ⇒ fof^-1: B → B

Clearly I_B : B → B such that I_B (y) = y ∀ y ∈ B
Let y ∈ B
y ∈ B, f^-1: B → A = f^-1(y) ∈ A
Let f^-1(y) = x
f^-1(y) = x ⇒ f (x) = y
(fof’)(y) = f [ f ^-1( y)] = f (x) = y = I_B (y)
∴ (fof^-1)(y) = I_B (y) ∀ y ∈ B
∴ fof^-1 = I_B

Theorem: If f : A → B, g : B → C are two bijections then (gof )^-1 = f^-1og^-1.
Proof:
f : A → B, g : B →C are bijections gof: A → C is bijection (gof )^-1: C → A is a bijection.
f : A → B is a bijection f^-1: B → A is a bijection
g : B → C Is a bijection ⇒ g^-1: C → B is a bijection
g^-1:C → B , g^-1: B → A are bijections ⇒ f^-1 og^-1: C → A is a bijection

Let z ∈ C
z ∈ C, g : B → C is onto ⇒ ∃ y ∈ B ∋ g (y) = z ⇒ g^-1(z) = y
y e B, f: A → B is onto ⇒ ∃ x ∈ A ∋ f (x) = y ⇒ f^-1(y) = x
(gof) (x) = g[ f (x)] = g (y) = z ⇒ (gof )^-1(z) = x
∴ (gof)^-1 (z) = x = f^-1( y) = f^-1 [ g^-1 (z) ] = (f ^-1og^-1)(z)
∴ (gof )^-1 = f^-1og^-1

Theorem:
If f : A → B, g : B → A are two functions such that gof = I_A and fog = I_B then f : A → B is a bijection and f^-1 = g .
Proof:
Let x₁, x₂ ∈ A, f (x₁) = f (x₂)
x₁, x₂ ∈ A, f : A → B ⇒ f (x₁), f (x₂) ∈ B
f (x₁), f (x₂) ∈ B, f (x₁) = f (x₂), g = B → A
⇒ g [ f (x₁)] = g[ f (x₂)]
⇒ (gof)(x₁) = (gof)(x₁) ⇒ I_A (x₂) ⇒ x₁ = x₂
x₁,x₂ ∈ A, f (x₁) = f (x₂) ⇒ x₁ = x₂.
∴ f : A → B is one one

Let y ∈ B .
y ∈ B, g : B → A ⇒ g(y) ∈ A

Def 11:
A function f : A → B is said tobe a constant function if the range of f contain only one element i.e., f (x) = c ∀ x ∈ A where c is a fixed element of B

Def 12:
A function f : A → B is said to be a real variable function if A ⊆ R.

Def 13:
A function f : A → B is said to be a real valued function iff B ⊆ R.

Def 14:
A function f : A → B is said to be a real function if A ⊆ R, B ⊆ R.

Def 15:
If f : A → R, g : B → R then f + g : A ∩ B → R is defined as (f + g)(x) = f (x) + g (x) ∀ x ∈ A ∩ B

Def 16:
If f : A → R and k e R then kf : A → R is defined as (kf)(x) = kf (x), ∀ x ∈ A

Def 17:
If f : A → B, g : B → R then fg : A n B → R is defined as (fg)(x) = f (x)g(x) ∀ x ∈ A ∩ B .

Def 18:
If f : A → R, g : B → R then : C → R is defined as
C = {x ∈ A n B: g(x) ≠ 0}.

Def 19:
If f : A → R then |f| (x) =| f (x)|, ∀ x ∈ A

Def 20:
If n ∈ Z , n ≥ 0, a₀, a₁, a₂, ………….. a_n ∈ R, a_n ≠ 0, then the function f : R → R defined by
f (x) = a₀ + a₁x + a₂x² + + a_nxⁿ ∀ x ∈ R is called a polynomial function of degree n.

Def 21:
If f : R → R, g : R → R are two polynomial functions, then the quotient f/g is called a rational function.

Def 22:
A function f : A → R is said to be bounded on A if there exists real numbers k1, k2 such that k1 < f (x) < k2 ∀ x ∈ A Def 23: A function f : A → R is said to be an even function if f (-x) = f (x) ∀ x ∈ A Def 24: A function f : A → R is said to be an odd function if f (-x) = – f (x) ∀ x ∈ A .  Def 25: If a ∈ R, a > 0 then the function f : R → R defined as f (x) = ax is called an exponential function.

Def 26:
If a ∈ R, a > 0, a ≠ 1 then the function f : (0, ∞) → R defined as f (x) = log_a x is called a logarithmic function.

Def 27:
The function f : R → R defined as f(x) = n where n ∈ Z such that n ≤ x < n + 1 ∀ x ∈ R is called step function or greatest integer function. It is denoted by f (x) = [x]

Def 28:
The functions f(x) = sin x, cos x, tan x, cot x, sec x or cosec x are called trigonometric functions.

Def 29:
The functions f (x) = sin^-1 x ,cos^-1x,tan^-1 x,cot^-1x,sec^-1x or cos ec^-1 x are called inverse trigonometric functions.

Def 30:
The functions f(x) = sinh x, cosh x, coth x, sech x or cosech x are called hyperbolic functions.

Def 30:
The functions f(x) = sinh^-1x, cosh^-1x, coth^-1x, sech^-1x or cosech^-1x are called Inverse hyperbolic functions.

Function	Domain	Range
ax	R	(0, ∞)
log a x	(0, ∞)	R
[X]	R	Z
[X]	R	[0, ∞)
√ x	[0, ∞)	[0, ∞)
sin x	R	[-1, 1]
cos x	R	[-1, 1]
tan x	R – {(2n +1)\(\frac{\pi}{2}\): n ∈ Z}	R
cot x	R – {nπ: n ∈ Z}	R
sec x	R – {(2n +1)\(\frac{\pi}{2}\): n ∈ Z}	(-∞,-1] ∪ [1, ∞)
cos ecx	R – {nπ: n ∈ Z}	(-∞,-1] ∪ [1, ∞)
Sin-1x	[-1 , 1]	[-π/2, π/2]
Cos-1x	[ -1, 1]	[0, π]
Tan-1x	R	(-π/2, π/2)
Cot-1x	R	(0, π)
Sec-1x	(-∞ -1] ∪ [1, ∞)	[0, π/2) ∪ (π/2, π]
Cosec-1x	(-∞ -1] ∪ [1, ∞)	[-π/2,0) ∪ (0, π/2]
sinh x	R	R
cosh x	R	[1, ∞)
tanh x	R	(-1,1)
coth x	(-∞,0) ∪ (0, ∞)	(-∞,-1) ∪ (1, ∞)
sech x	R	(0, 1]
cosech x	(-∞,0) ∪ (0, ∞)	(-∞,0) ∪ (0, ∞)
Sinh-1x	R	R
Cosh-1x	[1, ∞)	[0, ∞)
Tanh-1x	(-1, 1)	R
Coth-1x	(-∞,-1) ∪ (1, ∞)	(-∞,0) ∪ (0, ∞)
Sech-1x	(0, 1]	[0, ∞)
Coseh-1x	(-∞,0) ∪ (0, ∞)	(-∞,0) ∪ (0, ∞)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)

All problems in this exercise have reference to ΔABC.

I.

Question 1.
Express \(\Sigma r_{1} \cot \frac{A}{2}\) in terms of s.
Solution:
\(\Sigma r_{1} \cot \frac{A}{2}\) = \(\Sigma\left(s \tan \frac{A}{2}\right) \cot \frac{A}{2}\)
= Σs
= s + s + s
= 3s

Question 2.
Show that Σa cot A = 2(R + r).
Solution:
L.H.S = Σa . cot A
= Σ2R sin A \(\frac{\cos A}{\sin A}\)
= 2R Σ cos A
= \(2 R\left(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)\) (From transformants)
= \(2\left(R+4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)\)
= 2(R + r)
= R.H.S

Question 3.
In ∆ABC, prove that r₁ + r₂ + r₃ – r = 4R.
Solution:

Question 4.
In ∆ABC, prove that r + r₁ + r₂ – r₃ = 4R cos C.
Solution:

Question 5.
If r + r₁ + r₂ + r₃ then show that C = 90°.
Solution:

II.

Question 1.
Prove that 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)²
Solution:

Question 2.
Prove that \(\left(\frac{1}{r}-\frac{1}{r_{1}}\right)\left(\frac{1}{r}-\frac{1}{r_{2}}\right)\left(\frac{1}{r}-\frac{1}{r_{3}}\right)=\frac{a b c}{\Delta^{3}}=\frac{4 R}{r^{2} s^{2}}\)
Solution:

Question 3.
Prove that r(r₁ + r₂ + r₃) = ab + bc + ca – s².
Solution:

Question 4.
Show that \(\sum \frac{r_{1}}{(s-b)(s-c)}=\frac{3}{r}\)
Solution:

Question 5.
Show that \(\left(r_{1}+r_{2}\right) \tan \frac{C}{2}=\left(r_{3}-r\right) \cot \frac{C}{2}=c\)
Solution:

Question 6.
Show that r₁r₂r₃ = \(r^{3} \cot ^{2} \frac{A}{2} \cdot \cot ^{2} \frac{B}{2} \cdot \cot ^{2} \frac{C}{2}\)
Solution:

III.

Question 1.
Show that cos A + cos B + cos C = 1 + \(\frac{r}{R}\)
Solution:
L.H.S = cos A + cos B + cos C
= 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\)) + cos C

Question 2.
Show that \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=2+\frac{r}{2 R}\)
Solution:

Question 3.
Show that \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-\frac{r}{2 R}\)
Solution:

Question 4.
Show that
(i) a = (r₂ + r₃) \(\sqrt{\frac{r r_{1}}{r_{2} r_{3}}}\)
(ii) ∆ = r₁r₂ \(\sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}\)
Solution:

Question 5.
Prove that \(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r^{2}\) = 16R² – (a² + b² + c²).
Solution:

Question 6.
If p₁, p₂, p₃ are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that
(i) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=\frac{1}{r}\)
(ii) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}=\frac{1}{r_{3}}\)
(iii) p₁ . p₂ . p₃ = \(\frac{(a b c)^{2}}{8 R^{3}}=\frac{8 \Delta^{3}}{a b c}\)
Solution:

Question 7.
If a = 13, b = 14, c = 15, show that R = \(\frac{65}{8}\), r = 4, r₁ = \(\frac{21}{2}\), r₂ = 12 and r₃ = 14.
Solution:
a = 13, b = 14, c = 15
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+14+15}{2}\)
= 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6

Question 8.
If r₁ = 2, r₂ = 3, r₃ = 6 and r = 1, prove that a = 3, b = 4 and c = 5.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a)

All problems in this exercise refer to ΔABC

I.

Question 1.
Show that Σa(sin B – sin C) = 0
Solution:

Question 2.
If a = √3 + 1 cms, ∠B = 30°, ∠C = 45°, then find c.
Solution:
∠B = 30°, ∠C = 45° and a = (√3 + 1) cms
A = 180° – (B + C)
= 180° – (30° + 45°)
= 180° – 75°
= 105°

Question 3.
If a = 2 cms, b = 3 cms, c = 4 cms, then find cos A.
Solution:
a = 2 cms, b = 3 cms and c = 4 cms

Question 4.
If a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\), then find c.
Solution:
a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\)
c² = a² + b² – 2ab cos C
⇒ c² = 676 + 900 – 2 × 26 × 30 × \(\frac{63}{65}\)
⇒ c² = 1576 – 1512
⇒ c² = 64
⇒ c = 8

Question 5.
If the angles are in the ratio 1 : 5 : 6, then find the ratio of its sides.
Solution:
Given \(\frac{A}{1}=\frac{B}{5}=\frac{C}{6}\), B = 5A, C = 6A
A + B + C = 180°
⇒ A + 5A + 6A = 180°
⇒ 12A = 180°
⇒ A = 15°
∴ B = 5A = 75°
∴ C = 6A = 90°
a : b : c = sin A : sin B : sin C
= sin 15° : sin 75° : sin 90°
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}: \frac{\sqrt{3}+1}{2 \sqrt{2}}: 1\)
= (√3 – 1) : (√3 + 1) : 2√2

Question 6.
Prove that 2(bc cos A + ca cos B + ab cos C) = a² + b² + c².
Solution:
L.H.S = Σ2bc cos A
= Σ2bc \(\frac{\left(b^{2}+c^{2}-a^{2}\right)}{2 b c}\)
= Σ(b² + c² – a²)
= b² + c² – a² + c² + a² – b² + a² + b² – c²
= a² + b² + c²
= R.H.S

Question 7.
Prove that \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
Solution:
L.H.S = \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
[∵ c² = a² + b² – 2ab cos C and b² = a² + c² – 2ac cos B]

Question 8.
Prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S = (b + c) cos A + (c + a) cos B + (a + b) cos C
= (b cos A + c cos A) + (c cos B + a cos B) + (a cos C + b cos C)
= (b cos C + c cos B) + (a cos C + c cos A) + (a cos B + b cos A)
= a + b + c
= R.H.S

Question 9.
Prove that (b – a cos C) sin A = a cos A sin C
Solution:
LHS = (b – a cos C) sin A
= (a cos C + c cos A – a cos C) sin A
= c cos A sin A [∵ b = a cos C + c cos A]
= (2R sin C) cos A sin A
= a cos A sin C (∵ 2R sin A = a)

Question 10.
If 4, 5 are two sides of a triangle and the included angle is 60°, find its area.
Solution:
Let a = 4, b = 5 are two sides and included angle is C = 60° then
Area of ∆ = \(\frac{1}{2}\) ab sin C
= \(\frac{1}{2}\) × 4 × 5 × sin 60°
= 2 × 5 × \(\frac{\sqrt{3}}{2}\)
= 5√3 sq.cm

Question 11.
Show that \(b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=s\)
Solution:

Question 12.
If \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\), then show that ∆ABC is equilateral.
Solution:
Given that \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\)
⇒ \(\frac{2 R \sin A}{\cos A}=\frac{2 R \sin B}{\cos B}=\frac{2 R \sin C}{\cos C}\)
⇒ \(\frac{\sin A}{\cos A}=\frac{\sin B}{\cos B}=\frac{\sin C}{\cos C}\)
⇒ tan A = tan B = tan C
⇒ A = B = C
⇒ ∆ABC is an equilateral triangle.

II.

Question 1.
Prove that a cos A + b cos B + c cos C = 4R sin A sin B sin C.
Solution:
L.H.S = (2R sin A) cos A + (2R sin B) cos B + (2R sin C) cos C
= R (sin 2A + sin 2B + sin 2C)
= R (2 sin (A + B) cos (A – B) + sin 2C)
= R [2 sin (180° – C) cos (A – B) + sin 2C]
= R (2 sin C . cos (A – B) + 2 sin C . cos C)
= 2R sin C (cos (A – B)) + cos C)
= 2R sin C (cos (A – B) + cos (180° – \(\overline{\mathrm{A}+\mathrm{B}}\))
= 2R sin C [cos (A – B) – cos (A + B)]
= 2R sin C (2 sin A sin B)
= 4R sin A sin B sin C
= R.H.S.

Question 2.
Prove that Σa³ sin(B – C) = 0.
Solution:
L.H.S = Σa² [a sin (B – C)]
= Σa² [2R . sin A sin (B – C)]
= R Σ a² (2 sin (180° – \(\overline{\mathrm{B}+\mathrm{C}}\)) sin (B – C))
= R Σ a² [2 sin (B + C) . sin (B – C)]
= R Σ a² (sin² B – sin² C)
= R Σ a² \(\left(\frac{b^{2}}{4 R^{2}}-\frac{c^{2}}{4 R^{2}}\right)\)
= \(\frac{1}{2 R}\) Σ[a² (b² – c²)]
= \(\frac{1}{2 R}\) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)]
= \(\frac{1}{2 R}\) (a²b² – a²c² + b²c² – a²b² + a²c² – b²c²)
= \(\frac{1}{2 R}\) × 0
= 0
= R.H.S

Question 3.
Prove that \(\frac{a \sin (B-C)}{b^{2}-c^{2}}=\frac{b \sin (C-A)}{c^{2}-a^{2}}=\frac{c \sin (A-B)}{a^{2}-b^{2}}\)
Solution:

Question 4.
Prove that \(\sum a^{2} \frac{a^{2} \sin (B-C)}{\sin B+\sin C}=0\)
Solution:

Question 5.
Prove that \(\frac{a}{b c}+\frac{\cos A}{a}=\frac{b}{c a}+\frac{\cos B}{b}=\frac{c}{a b}+\frac{\cos C}{c}\)
Solution:

Question 6.
Prove that \(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}\)
Solution:

Question 7.
If C = 60°, then show that
(i) \(\frac{\mathbf{a}}{\mathbf{b}+\mathbf{c}}+\frac{\mathbf{b}}{\mathbf{c}+\mathbf{a}}=1\)
(ii) \(\frac{b}{c^{2}-a^{2}}+\frac{a}{c^{2}-b^{2}}=0\)
Solution:
∠C = 60°
⇒ c² = a² + b² – 2ab cos C
⇒ c² = a² + b² – 2ab cos 60°
⇒ c² = a² + b² – 2ab (\(\frac{1}{2}\))
⇒ c² = a² + b² – ab

Question 8.
If a : b : c = 7 : 8 : 9, find cos A : cos B : cos C.
Solution:

Question 9.
Show that \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\)
Solution:

Question 10.
Prove that (b – a) cos C + c (cos B – cos A) = c . sin(\(\frac{A-B}{2}\)) cosec(\(\frac{A+B}{2}\))
Solution:
L.H.S = b cos C – a cos C + c cos B – c cos A
= (b cos C + c cos B) – (a cos C + c cos A)
= a – b
= 2R (sin A – sin B)

Question 11.
Express \(a \sin ^{2} \frac{C}{2}+c \cdot \sin ^{2} \frac{A}{2}\) interms of s, a, b, c.
Solution:

Question 12.
If b + c = 3a, then rind the value of \(\cot \frac{\mathrm{B}}{2} \cot \frac{\mathrm{C}}{2}\)
Solution:

Question 13.
Prove that (b + c) cos (\(\frac{B+C}{2}\)) = a cos (\(\frac{B-C}{2}\))
Solution:

Question 14.
In ΔABC, show that \(\frac{b^{2}-c^{2}}{a^{2}}=\sin \frac{(B-C)}{(B+C)}\)
Solution:

III.

Question 1.
Prove that
(i) \(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{s^{2}}{\Delta}\)
(ii) \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\) = \(\frac{b c+c a+a b-s^{2}}{\Delta}\)
(iii) \(\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}\) = \(\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\)
Solution:

Question 2.
Show that
(i) Σ(a + b) tan(\(\frac{A-B}{2}\)) = 0
(ii) \(\frac{\mathbf{b}-\boldsymbol{c}}{\mathbf{b}+\mathbf{c}} \cot \frac{A}{2}+\frac{b+c}{b-c} \tan \frac{A}{2}\) = 2 cosec(B – C)
Solution:

Question 3.
(i) If sin θ = \(\frac{a}{b+c}\), then show that cos θ = \(\frac{2 \sqrt{b c}}{b+c} \cdot \cos \frac{A}{2}\)
Solution:

(ii) If a = (b + c) cos θ, then prove that sin θ = \(\frac{2 \sqrt{b c}}{b+c} \cos \left(\frac{A}{2}\right)\)
Solution:

(iii) For any anlge θ show that a cos θ = b cos (C + θ) + c cos (B – θ).
Solution:
b cos (C + θ) + c cos (B – θ)
= b (cos C . cos θ – sin C sin θ) + c (cos B cos θ + sin B sin θ)
= (b cos C + c cos B) cos θ + (-b sin C + C sin B) sin θ
= a cos θ + (-2R sin B sin C + 2R sin B sin C) sin θ
= a cos θ

Question 4.
If the angles of ∆ABC are in A.P and b : c = √3 : √2 , then show that A = 75°.
Solution:
∵ The angles A, B, C of a triangle are in A.P.
⇒ 2B = A + C
⇒ 3B = A + B + C
⇒ 3B = 180°
⇒ B = 60°

Question 5.
If \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=\frac{\sin C}{\sin (A-B)}\), prove that ∆ABC is either isosceles or right-angled.
Solution:

⇒ sin 2A = sin 2B
⇒ A = B
⇒ ∆ABC is isosceles
or 2A = 180° – 2B
or A = 90° – B
or A + B = 90°
so A ≠ B ⇒ ∆ABC is a right-angled triangle
∴ ∆ABC is either isosceles (or) right-angled.

Question 6.
If cos A + cos B + cos C = \(\frac{3}{2}\), then show that the triangle is equilateral.
Solution:

Question 7.
If cos² A + cos² B + cos² C = 1, then show that ∆ABC is right-angled.
Solution:
Given cos² A + cos² B + cos² C = 1 …….(1)
Now cos² A + cos² B + cos² C
= cos² A + 1 – sin² B + cos² C
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) cos (A – B) + cos² C
= 1 + cos (180° – C) cos (A – B) + cos² C
= 1 – cos C . cos (A – B) + cos² C
= 1 – cos C (cos (A – B) – cos C)
= 1 – cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 1 – cos C (cos (A – B) + cos (A + B))
= 1 – cos C (2 cos A cos B)
= 1 – 2 cos A cos B cos C
Substituting in (1) we get
1 – 2 cos A cos B cos C = 1
∴ 2 cos A cos B cos C = 0
∴ cos A = 0 or cos B = 0 or cos C = 0
i.e., A = 90° or B = 90° or C = 90°
∴ ∆ABC is right-angled.

Question 8.
If a² + b² + c² = 8R², then prove that the triangle is right angled.
Solution:
Given a² + b² + c² = 8R²
⇒ 4R² (sin² A + sin² B + sin² C) = 8R²
⇒ sin² A + sin² B + sin² C = 2 ……(1)
Now sin² A + sin² B + sin² C
= 1 – cos² A + sin² B + sin² C
= 1 – (cos² A – sin² B) + sin² C
= 1 – cos (A + B) . cos (A – B) + sin² C
= 1 – cos (180° – C) cos (A – B) + sin² C
= 1 + cos C cos (A – B) + 1 – cos² C
= 2 + cos C (cos (A – B) – cos C)
= 2 + cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 2 + cos C (cos (A – B) + cos (A + B))
= 2 + cos C (2 cos A cos B)
= 2 + 2 cos A cos B cos C
Substituting in (1), we get
2 + 2 cos A cos B cos C = 2
2 cos A cos B cos C = 0
⇒ cos A = 0 or cos B = 0 or cos C = 0
∴ A = 90° or B = 90° or C = 90°
∴ ΔABC is right-angled.

Question 9.
If cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P., then prove that a, b, c are in A.P.
Solution:
∵ cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P
⇒ \(\frac{(s)(s-a)}{\Delta}, \frac{(s)(s-b)}{\Delta}, \frac{(s)(s-c)}{\Delta}\) are in A.P.
⇒ s – a, s – b, s – c are in A.P
⇒ -a, -b, -c are in A.P
⇒ a, b, c are in A.P

Question 10.
If \(\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}, \sin ^{2} \frac{C}{2}\) are in H.P., show that a, b, c are in H.P.
Solution:

Question 11.
If C = 90° then prove that \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\) sin (A – B) = 1
Solution:

Question 12.
Show that \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\) = ∆
Solution:
LHS = \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\)
= \(\frac{4 R^{2} \sin ^{2} A}{4}\) 2 sin C cos C + \(\frac{4 R^{2} \sin ^{2} C}{4}\) 2 sin A cos A
= 2R² sin²A sin C cos C + 2R² sin²C sin A cos A
= 2R² sin A sin C (sin A cos C + cos A sin C)
= 2R² sin A sin C sin(A + C)
= 2R² sin A sin C sin(180 – B)
= 2R² sin A sin B sin C
= ∆
= RHS

Question 13.
A lamp post is situated at the middle point I the side AC of a triangular plot A with BC = 7 meters, CA = 8 meters, and AB = 9 meters, lamp post subtends an angle of 15° at point B. find H height of the lamp post.
Solution:
Let MP be the height of the lamp post

Question 14.
Two ships leave a port at the same time. One goes 24 km. per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.
Solution:
Given the first ship goes 24 km per hour After 3 hours the first ship goes 72 km.

Given the second ship goes 32 km per hour After 3 hours, the second ship goes 96 km.
Let AB = x
∠AOB = 180° – (75 + 45) = 60°
Apply cosine rule for ∆AOB,
cos 60° = \(\frac{(72)^{2}+(96)^{2}-x^{2}}{2(72)(96)}\)
⇒ \(\frac{1}{2}=\frac{5184+9216-x^{2}}{13824}\)
⇒ 13824 = 28800 – 2x²
⇒ 2x² = 14976
⇒ x² = 7488
⇒ x = 86.4 (Appx)
At the end of 3 hours the difference between ships was 86.4 km.

Question 15.
A tree stands vertically on the slant of the hill. From A point on the ground 35 meters down the hill from the base of the tree, the angle, the elevation of the top of the tree is 60° if the angle of elevation of the foot of the tree from A is 15°, then find the height of the tree.
Solution:
Let BC be the height of the tree
BC = h
Let BD = x, AD = y
Given AB = 35 m

Question 16.
The upper \(\frac{3}{4}\)th portion of a vertical pole subtends an angle tan-1(\(\frac{3}{5}\)) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. Given that the vertical pole is at a height less than 100 m from the ground, find its height.
Solution:

⇒ 6400 + h² = 200h
⇒ h² – 200h + 6400 = 0
⇒ h² – 160h – 40h + 6400 = 0
⇒ h(h – 160) – 40(h – 160) = 0
⇒ (h – 160) (h – 40) = 0
⇒ h = 40 or 160
but the height of the pole should be less than 100m
∴ h = 40 m

Question 17.
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D, the angle of elevation of point A is 45°. Find the height of the pole.
Solution:
Let ‘h’ be the height of the pole
AB = h
given CD = 7 m
∠ACB = 60°, ∠ADB = 45°, Let BC = x

Question 18.
Let an object be placed at some height h cm and let P and Q two points of observation which are at a distance of 10 cm apart on a line inclined at an angle of 15° to the horizontal. If the angles of elevation of the object from P and Q are 30° and 60° respectively then find h.
Solution:
Let AB = Height of the object from ‘A’ = h m
Given that P & Q are two observations,
PQ = 10 cms

From ∆APB,
∠P = 30; ∠A = 90; ∠B = ?
A + P + B = 180°
⇒ 30° + 30° + B = 180°
⇒ B = 180° – 120°
⇒ B = 60°
From ∆BQC,
∠Q = 60°; ∠C = 90°; ∠B = ?
Q + C + B = 180°
⇒ 60° + 90° + B = 180°
⇒ B = 180° – 150°
⇒ B = 30°
From ∆BQP,
∠P = 15; ∠B = 30; ∠Q = ?
P + B + Q = 180°
⇒ 15° + 30° + Q = 180°
⇒ Q = 180° – 45°
⇒ Q = 135°
Applying the ‘sin’ rule for ∆BQP

From ∆PAB,
sin 30° = \(\frac{B A}{B P}\)
BP . sin 30° = AB = h
√2 × 10 × \(\frac{1}{2}\) = AB = h
5√2 = AB = h
∴ h = 5√2

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a)

Question 1.
If sinh x = \(\frac{3}{4}\), find cosh (2x) and sinh (2x).
Solution:

Question 2.
If sinh x = 3, then show that x = log_e(3 + √10).
Solution:

Question 3.
Prove that
(i) tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)
Solution:

(ii) coth (x – y) = \(\frac{{coth} x \cdot {coth} y-1}{{coth} y-{coth} x}\)
Solution:

Question 4.
Prove that
(i) (cosh x – sinh x)ⁿ = cosh (nx) – sinh (nx), for any n ∈ R.
Solution:

(ii) (cosh x + sinh x)ⁿ = cosh (nx) + sinh (nx), for any n ∈ R.
Solution:

Question 5.
Prove that \(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2 cosech x, for x ≠ 0
Solution:

Question 6.
Prove that \(\frac{\cosh x}{1-\tanh x}+\frac{\sinh x}{1-{coth} x}\) = sinh x + cosh x, for x ≠ 0
Solution:

Question 7.
For any x ∈ R, prove that cosh⁴x – sinh⁴x = cosh (2x)
Solution:
L.H.S = cosh⁴x – sinh⁴x
= (cosh²x)² – (sinh²x)²
= [cosh²x – sinh²x] [cosh²x + sinh²x]
= (1) cosh (2x)
= cosh (2x)
∴ cosh⁴x – sinh⁴x = cosh (2x)

Question 8.
If u = \(\log _{e}\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)\) and if cos θ > 0,then prove that cosh u = sec θ.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a)

I.

Question 1.
Evaluate the following.
(i) \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)
Solution:

(ii) \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
\(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\cos ^{-1}\left(\cos \frac{\pi}{4}\right)=\frac{\pi}{4}\)

(iii) sec^-1(-√2)
Solution:

(iv) cot^-1(-√3)
Solution:

(v) \(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\)
Solution:

(vi) \(\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)\)
Solution:

(vii) \(\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)
Solution:
\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)

(ii) \(\tan \left({cosec}^{-1} \frac{65}{63}\right)\)
Solution:
\(\tan \left({cosec}^{-1} \frac{65}{63}\right)=\tan \left(\tan ^{-1} \frac{63}{16}\right)\) = \(\frac{63}{16}\)

(iii) \(\sin \left(2 \sin ^{-1} \frac{4}{5}\right)\)
Solution:

(iv) \(\sin ^{-1}\left(\sin \frac{33 \pi}{7}\right)\)
Solution:

(v) \(\cos ^{-1}\left(\cos \frac{17 \pi}{6}\right)\)
Solution:

Question 3.
Simplify each of the following.
(i) \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)
Solution:

(ii) tan^-1(sec x + tan x)
Solution:

(iii) \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

(iv) sin^-1(2 cos²θ – 1) + cos-1(1 – 2 sin²θ)
Solution:
sin^-1(cos 2θ) + cos^-1(cos 2θ)
= sin^-1[sin (90° – 2θ)] + cos^-1(cos 2θ)
= 90° – 2θ + 2θ
= 90°

(v) \(\tan ^{-1}\left(x+\sqrt{1+x^{2}}\right)\); x ∈ R
Solution:

II.

Question 1.
Prove that
(i) \(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{36}{85}\)
Solution:

(ii) \(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\)
Solution:

(iii) \(\tan \left[\cot ^{-1} 9+{cosec}^{-1} \frac{\sqrt{41}}{4}\right]=1\)
Solution:

(iv) \(\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{\sqrt{34}}=\tan ^{-1} \frac{27}{11}\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}\right)\)
Solution:

(ii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{5}{\sqrt{34}}\right)\)
Solution:

(iii) \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)
Solution:
Let \(\sin ^{-1} \frac{3}{5}\) = α and \(\sin ^{-1} \frac{5}{13}\) = β

Question 3.
Prove that
(i) \(\cos \left[2 \tan ^{-1} \frac{1}{7}\right]=\sin \left[2 \tan ^{-1} \frac{3}{4}\right]\)
Solution:

(ii) \(\tan \left[2 \tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right]=2\)
Solution:

(iii) \(\cos \left\{2\left[\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\right]\right\}=\frac{3}{5}\)
Solution:

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}-\tan ^{-1} \frac{2}{9}=0\)
Solution:

(ii) \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

(iii) \(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}=\frac{\pi}{4}\)
Solution:

(iv) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\) = \(\cot ^{-1} \frac{201}{43}+\cot ^{-1} 18\)
Solution:

Question 5.
Show that
(i) sec² (tan^-1 2) + cosec² (cot^-1 2) = 10
Solution:
Let a = tan^-1 2 ⇒ tan α = 2
sec² α = 1 + tan^-1 α = 1 + 4 = 5
Let β = cot^-1 2 ⇒ cot β = 2
cosec² β = 1 + cot² β = 1 + 4 = 5
LHS = sec² (tan^-1 2) + cosec² (cot^-1 2)
= 5 + 5
= 10
= RHS

(ii) Find the value of \(\left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)\)
Solution:

(iii) If sin^-1 x – cos^-1 x = \(\frac{\pi}{6}\) then find x.
Solution:

III.

Question 1.
Prove that
(i) \(2 \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{5}{13}=\cos ^{-1} \frac{323}{325}\)
Solution:

(ii) \(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}\)
Solution:

(iii) \(4 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{99}-\tan ^{-1} \frac{1}{70}=\frac{\pi}{4}\)
Solution:

Question 2.
(i) If α = \({tan}^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)\) then prove that x² = sin 2α.
Solution:

(ii) Prove that tan\(\left\{2-\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\right\}=\mathbf{x}\)
Solution:

(iii) Prove that \(\sin \left[\cot ^{-1} \frac{2 x}{1-x^{2}}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]\) = 1
Solution:

(iv) Prove that \(\left\{\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right\}\)
Solution:

Question 3.
(i) If cos^-1 p + cos^-1 q + cos^-1 r = π, then prove that p² + q² + r² + 2pqr = 1
Solution:
Let cos^-1 p = A, cos^-1 q = B and cos^-1 r = C
then A + B + C = π ………(1)
and p = cos A, q = cos B and r cos C
Now p² + q² + r² = cos² A + cos² B + cos² C
= cos² A + (1 – sin² B + cos² C)
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) . cos (A – B) + cos² C
= 1 + cos (π – C) cos (A – B) + cos² C (By (1))
= 1 – cos C cos (A – B) + cos² C
= 1 – cos C [cos (A – B) – cos C]
= 1 – cos C [cos (A – B) – cos(180° – \(\overline{A+B}\)]
= 1 – cos C [cos (A – B) + cos (A + B)]
= 1 – cos C [2 cos A cos B]
= 1 – 2 pqr
∴ p² + q² + r² + 2pqr = 1

(ii) If \(\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right)\) = \(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\), then prove that x = \(\frac{p-q}{1+p q}\)
Solution:

(iii) If a, b, c are distinct non-zero real numbers having the same sign, prove that \(\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)\) + \(\cot ^{-1}\left(\frac{c a+1}{c-a}\right)\) = π or 2π
Solution:
Since (a – b) + (b – c) + (c – a) = 0.
(a – b), (b – c), (c – a) all cannot have the same sign.
Now two cases arise, namely, either two of these numbers are positive and one negative (or) two of these numbers are negative and one is positive.
Case (i): Without loss of generality, we assume that (a – b), (b – c) are both positive and (c – a) is negative

Case (ii): Without loss of generality, we assume that (a – b) and (b – c) are both negative and (c – a) is positive.

(iv) If sin^-1 (x) + sin^-1 (y) + sin^-1 (z) = π, then prove that \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)
Solution:
Let sin^-1 (x) = A, sin^-1 (y) = B and sin^-1 (z) = C
Then A + B + C = π …………(1)
and x = sin A, y = sin B and z = sin C
Now LHS = \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}\)
= sin A \(\sqrt{1-\sin ^{2} A}\) + sin B \(\sqrt{1-\sin ^{2} B}\) + sin C \(\sqrt{1-\sin ^{2} C}\)
= sin A cos A + sin B cos B + sin C cos C
= \(\frac{1}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{1}{2}\) [2 . sin (A + B) cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin (π – c). cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin C cos (A – B) + 2 sin C cos C]
= \(\frac{1}{2}\) (2 sin C) [cos (A – B) + cos C]
= sin C [cos (A – B) + cos (180° – \(\overline{A+B}\)]
= sin C [cos (A – B) – cos (A + B)]
= sin C [2 sin A sin B]
= 2 xyz
∴ \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)

(v) (a) If tan^-1 x + tan^-1 y + tan^-1 z = π, then prove that x + y + z = xyz.
Solution:
Let A = tan^-1 x, B = tan^-1 y, C = tan^-1 z
tan A = x, tan B = y, tan C = z
Given A + B + C = π ……….(1)
A + B = π – C
tan (A + B) = tan (π – C)
\(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
\(\frac{x+y}{1-x y}\) = -z
x + y = -z + xyz
∴ x + y + z = xyz

(b) If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\), then prove that xy + yz + zx = 1.
Solution:
Let tan^-1 x = A, tan^-1 y = B and tan^-1 z = C
Then A + B + C = \(\frac{\pi}{2}\) …….(1)
and x = tan A, y = tan B and z = tan C
∵ A + B + C = \(\frac{\pi}{2}\)
A + B = \(\frac{\pi}{2}\) – C
⇒ tan (A + B) = tan(\(\frac{\pi}{2}\) – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\) = cot C
⇒ \(\frac{x+y}{1-x y}=\frac{1}{z}\)
⇒ zx + yz = 1 – xy (or) xy + yz + zx = 1

Question 4.
Solve the following equations for x:
(i) \({Tan}^{-1}\left(\frac{x-1}{x-2}\right)+{Tan}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
Solution:

(ii) \(\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)\) = \(\tan ^{-1} \frac{2}{x^{2}}\)
Solution:
Given that

⇒ x²(3x + 1) = 2x(4x + 3)
⇒ x [x(3x + 1) – 2(4x + 3)] = 0
⇒ x = 0 (or) 3x² – 7x – 6 = 0
⇒ x = 0 (or) 3x² – 9x + 2x – 6 = 0
⇒ x = 0 (or) 3x(x – 3) + 2(x – 3) = 0
⇒ x = 0 (or) (3x + 2) (x – 3) = 0
⇒ x = 0 (or) 3 (or) \(\frac{-2}{3}\)

(iii) \(3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) + \(2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}\)
Solution:

(iv) sin^-1(1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
Given sin^-1(1 – x) – 2 sin^-1x = \(\frac{\pi}{2}\)
Let sin^-1(1 – x) = α and sin^-1(x) = β
Then sin α = 1 – x and sin β = x
cos α = \(\sqrt{1-(1-x)^{2}}\) and cos β = \(\sqrt{1-x^{2}}\)
Now sin^-1(1 – x) – 2 sin^-1(x) = \(\frac{\pi}{2}\)
⇒ α – 2β = \(\frac{\pi}{2}\)
⇒ α = \(\frac{\pi}{2}\) + 2β
⇒ sin α = sin (\(\frac{\pi}{2}\) + 2β)
⇒ sin α = cos 2β
⇒ 1 – x = 1 – 2 sin2β
⇒ 1 – x = 1 – 2x²
⇒ 2x² – x = 0
⇒ x(2x – 1) = 0
⇒ x = 0 (or) x = \(\frac{1}{2}\)
But when x = \(\frac{1}{2}\)

Hence x = 0 is the only solution for the given equation.

Question 5.
Solve the following equations.
(i) \(\cot ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{1}{2} \cot ^{-1}\left(\frac{1}{x}\right)\), x > 0 and x ≠ 1
Solution:

(ii) \(\tan \left[\cos ^{-1} \frac{1}{x}\right]=\sin \left[\cot ^{-1} \frac{1}{2}\right]\); x ≠ 0
Solution:
Let \(\cos ^{-1}\left(\frac{1}{x}\right)=\alpha, \cot ^{-1} \frac{1}{2}=\beta\)

(iii) \(\cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\frac{\pi}{6}\)
Solution:

(iv) \(\cos ^{-1}(\sqrt{3} \cdot x)+\cos ^{-1} x=\frac{\pi}{2}\)
Solution:

(v) \(\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Equations Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Equations Solutions Exercise 7(a)

I.

Question 1.
Find the principal solutions of the angles in the equations
(i) 2 cos²θ = 1
Solution:
cos²θ = \(\frac{1}{2}\)
θ = 45°, 135°

(ii) √3 sec θ + 2 = 0
Solution:
sec θ = \(\frac{-2}{\sqrt{3}}\)
⇒ cos θ = \(\frac{-\sqrt{3}}{2}\)
⇒ θ = 150°

(iii) 3 tan²θ = 1
Solution:
tan²θ = \(\frac{1}{3}\)
θ = ±\(\frac{\pi}{6}\)

Question 2.
Solve the following equations.
(i) cos 2θ = \(\frac{\sqrt{5}+1}{4}\), θ ∈ [0, 2π]
Solution:
cos 2θ = \(\frac{\sqrt{5}+1}{4}\)
⇒ cos 2θ = cos 36° = cos (\(\frac{\pi}{5}\))
and \(\frac{\pi}{5}\) ∈ [0, 2π]
2θ = \(\frac{\pi}{5}\) ⇒ θ = \(\frac{\pi}{10}\) is the principal solution
and 2θ = 2nπ ± \(\frac{\pi}{5}\) where n ∈ Z is the general solution
⇒ θ = nπ ± \(\frac{\pi}{10}\)
The values of θ in [0, 2π] are \(\left\{\frac{\pi}{10}, \frac{9 \pi}{10}, \frac{11 \pi}{10}, \frac{19 \pi}{10}\right\}\)

(ii) tan²θ = 1, θ ∈ [-π, π]
Solution:
tan²θ = 1 ⇒ tan θ = ±1
tan θ = ±1 = tan (\(\pm \frac{\pi}{4}\))
The principal solutions are θ = \(\pm \frac{\pi}{4}\)
and the general solution is given by nπ ± \(\frac{\pi}{4}\), n ∈ Z
Put n = -1, 0, 1
\(\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}\right\}\) is the solution set for the given equation in [-π, π]

(iii) sin 3θ = \(\frac{\sqrt{3}}{2}\), θ ∈ [-π, π]
Solution:
sin 3θ = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
and \(\frac{\pi}{3}\) ∈ [-π, π]
∴ 3θ = \(\frac{\pi}{3}\) is the principal solution
and 3θ = \(n \pi+(-1)^{n} \frac{\pi}{3}\), n ∈ Z
⇒ θ = \(\frac{n \pi}{3}+(-1)^{n} \cdot \frac{\pi}{9}\), n ∈ Z is the general solution.
The solution of θ in [-π, π] are \(\left\{\frac{-5 \pi}{9}, \frac{-4 \pi}{9}, \frac{\pi}{9}, \frac{2 \pi}{9}, \frac{7 \pi}{9}, \frac{8 \pi}{9}\right\}\)

(iv) cos²θ = \(\frac{3}{4}\), θ ∈ [0, π]
Solution:
cos²θ = \(\frac{3}{4}\)
⇒ cos θ = ±\(\frac{\sqrt{3}}{4}\)
The general solution is given by
θ = nπ ± \(\frac{\pi}{6}\), n ∈ Z
Put n = 0, 1
The solution set for the given equation in [0, π] is \(\left\{\frac{\pi}{6}, \frac{5 \pi}{6}\right\}\)

(v) 2 sin²θ = sin θ, θ ∈ (0, π)
Solution:
2 sin²θ – sin θ = 0
sin θ (2 sin θ – 1) = 0
sin θ = 0 or sin θ = \(\frac{1}{2}\)
since θ ∈ (0, π)
∴ The solution of θ = \(\left\{\frac{\pi}{6}+\frac{5 \pi}{6}\right\}\)

Question 3.
Find general solutions to the following equations.
(i) sin θ = \(\frac{\sqrt{3}}{2}\), cos θ = \(\frac{-1}{2}\)
Solution:
sin θ = \(\frac{\sqrt{3}}{2}\), cos θ = \(\frac{-1}{2}\)
∵ sin θ is +ve and cos θ is -ve
⇒ θ lies in II quadrant
sin θ = \(\frac{\sqrt{3}}{2}=\sin \left(\frac{2 \pi}{3}\right)\)
cos θ = \(\frac{-1}{2}=\cos \left(\frac{2 \pi}{3}\right)\)
⇒ θ = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ Z is the general solution.

(ii) tan x = \(\frac{-1}{\sqrt{3}}\), sec x = \(\frac{2}{\sqrt{3}}\)
Solution:
∵ tan x = \(\frac{-1}{\sqrt{3}}\) and sec x = \(\frac{2}{\sqrt{3}}\)
⇒ x lies in IV quadrant
tan x = \(\frac{-1}{\sqrt{3}}=\tan \left(\frac{-\pi}{6}\right)\)
sec x = \(\frac{2}{\sqrt{3}}=\sec \left(\frac{-\pi}{6}\right)\)
∴ θ = 2nπ + \(\left(\frac{-\pi}{6}\right)\), n ∈ Z is the general solution.

(iii) cosec θ = -2, cot θ = -√3
Solution:
cosec θ = -2, cot θ = -√3
⇒ θ lies in IV quadrant
cosec θ = -2
⇒ sin θ = \(-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right)\)
cot θ = -√3
⇒ tan θ = \(-\frac{1}{\sqrt{3}}=\tan \left(-\frac{\pi}{6}\right)\)
∴ θ = 2nπ + \(\left(\frac{-\pi}{6}\right)\), n ∈ Z is the general solution.

Question 4.
(i) If sin (270° – x) = cos 292°, then find x in (0, 360°).
Solution:
sin (270° – x) = cos (292°)
⇒ -cos x = cos (180° + 112°)
⇒ -cos x = -cos 112°
⇒ cos x = cos 112°
⇒ x = 112° or x = 360° – 112° = 248°

(ii) If x < 90°and sin (x + 28°) = cos (3x – 78°), then find x.
Solution:
sin (x + 28°) = cos (3x – 78°)
= sin (90° – 3x + 78°)
= sin (168° – 3x)
x + 28° = 168° – 3x + 28° (180°) or
= 180° – (168° – 3x) + 2x (180°)
⇔ there exists n ∈ Z such that
4x = 140° + 2x (180°)
2x = 16° – 2x (180°)
⇔ there exists n ∈ z such that
x = 35° + x(90°) or x = 8° – x (180°)
Hence x = 8° and x = 35° are the only values of x that lie (0, 90°) and satisfy the given equation.

Question 5.
Find general solutions to the following equations.
(i) 2 sin²θ = 3 cos θ
Solution:
2 sin²θ = 3 cos θ
⇒ 2(1 – cos²θ) = 3 cos θ
⇒ 2 cos²θ + 3 cos θ – 2 = 0
⇒ 2 cos²θ + 4 cos θ – cos θ – 2 = 0
⇒ 2 cos θ (cos θ + 2) – 1 (cos θ + 2) = 0
⇒ (2 cos θ – 1) (cos θ + 2) = 0
⇒ cos θ = \(\frac{1}{2}\) (or) cos θ = -2
∴ The range of cos θ is [-1, 1]
cos θ = -2 is not admissible
∴ cos θ = \(\frac{1}{2}=\cos \frac{\pi}{3}\)
⇒ θ = \(\frac{\pi}{3}\) is the principal solution and
θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.

(ii) sin²θ – cos θ = \(\frac{1}{4}\)
Solution:
sin²θ – cos θ = \(\frac{1}{4}\)
⇒ 4(1 – cos²θ) – 4 cos θ = 1
⇒ 4 cos²θ + 4 cos θ – 3 = 0
⇒ 4 cos²θ + 6 cos θ – 2 cos θ – 3 = 0
⇒ 2 cos θ (2 cos θ + 3) – (2 cos θ + 3) = 0
⇒ (2 cos θ – 1) (2 cos θ + 3) = 0
∴ cos θ = \(\frac{1}{2}\) (or) cos θ = \(\frac{-3}{2}\)
∵ The range of cos θ is [-1, 1]
cos θ = \(\frac{-3}{2}\) is not admisable
∴ cos θ = \(\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)\)
θ = \(\frac{\pi}{3}\) is the principal solution and
θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.

(iii) 5 cos²θ + 7 sin²θ = 6
Solution:
5 cos²θ + 7 sin²θ = 6
Dividing by cos²θ
⇒ 5 + 7 tan²θ = 6 sec²θ
⇒ 5 + 7 tan²θ = 6(1 + tan²θ)
⇒ tan²θ = 1
⇒ tan θ = ±1
∴ θ = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution.

(iv) 3 sin⁴x + cos⁴x = 1
Solution:
3 sin⁴x + cos⁴x = 1
⇒ 3 sin⁴x + (cos²x)² = 1
⇒ 3 sin⁴x + (1 – sin²x)² = 1
⇒ 3 sin⁴x + 1 + sin⁴x – 2 sin²x = 1
⇒ 4 sin⁴x – 2 sin²x = 0
⇒ 2 sin²x (2 sin²x – 1) = 0
⇒ sin x = 0 (or) sin x = ±\(\frac{1}{\sqrt{2}}\)
If sin x = 0
⇔ x = nπ, n ∈ Z is the general solution.
If sin x = ±\(\frac{1}{\sqrt{2}}\)
⇒ x = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution.
∴ General solution is x = nπ (or) nπ ± \(\frac{\pi}{4}\), n ∈ Z.

II.

Question 1.
Solve the following equations and write a general solution.
(i) 2 sin²θ – 4 = 5 cos θ
Solution:
2(1 – cos²θ) – 4 = 5 cos θ
2 – 2cos²θ – 4 = 5 cos θ
2 cos²θ + 5 cos θ + 2 = 0
2 cos²θ + 4 cos θ + cos θ + 2 = 0
2 cos θ (cos θ + 2) + 1 (cos θ + 2) = 0
(cos θ + 2) (2 cos θ + 1) = 0
cos θ = -2 or cos θ = \(-\frac{1}{2}\)
cos θ = -2 is not possible
∴ cos θ = \(-\frac{1}{2}\) = cos \(\frac{2 \pi}{3}\)
∴ General solution is θ = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ Z

(ii) 2 + √3 sec x – 4 cos x = 2√3
Solution:
2 + √3 sec x – 4 cos x = 2√3
\(\frac{2 \cos x+\sqrt{3}-4 \cos ^{2} x}{\cos x}\) = 2√3
2 cos x + √3 – 4 cos²x = 2√3 cos x
4 cos²x + 2√3 cos x – 2 cos x – √3 = 0
2 cos x (2 cos x + √3) – 1 (2 cos x + √3) = 0
(2 cos x – 1) (2 cos x + √3) = 0
cos x = \(\frac{1}{2}\) (or) cos x = \(\frac{-\sqrt{3}}{2}\)
If cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\), n ∈ Z
General solution x = 2nπ ± \(\frac{\pi}{3}\)
If cos x = \(\frac{-\sqrt{3}}{2}\) = cos \(\frac{5 \pi}{3}\)
General solution x = 2nπ ± \(\frac{5 \pi}{3}\), n ∈ Z

(iii) 2 cos²θ + 11 sin θ = 7
Solution:
2 (1 – sin²θ) + 11 sin θ = 7
2 – 2 sin²θ + 11 sin θ = 7
2 sin²θ – 11 sin θ + 5 = 0
2 sin²θ – 10 sin θ – sin θ + 5 = 0
2 sin θ (sin θ – 5) – 1(sin θ – 5) = 0
(sin θ – 5) (2 sin θ – 1) = 0
sin θ = 5 or sin θ = \(\frac{1}{2}\)
If sin θ = 5 is not possible
∴ sin θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
General solultion θ = nπ + (-1)ⁿ \(\frac{\pi}{6}\), n ∈ Z

(iv) 6 tan²x – 2 cos²x = cos 2x
Solution:
6 tan²x – 2 cos²x = cos 2x
⇒ 6(sec²x – 1) – 2 cos²x = 2 cos²x – 1
⇒ 6 sec²x – 6 – 4 cos²x + 1 = 0
⇒ 6 sec²x – 4 cos²x – 5 = 0
⇒ \(\frac{6}{\cos ^{2} x}\) – 4 cos²x – 5 = 0
⇒ 6 – 4 cos⁴x – 5 cos²x = 0
⇒ 4 cos⁴x + 5 cos²x – 6 = 0
⇒ 4 cos⁴x + 8 cos²x – 3 cos²x – 6 = 0
⇒ 4 cos²x (cos²x + 2) – 3(cos²x + 2) = 0
⇒ (4 cos²x – 3) (cos²x + 2) = 0
⇒ 4 cos²x = 3, cos²x ≠ -2
⇒ cos x = ±\(\frac{\sqrt{3}}{2}\)
∴ x = nπ ± \(\frac{\pi}{6}\), n ∈ Z is the general solution.

(v) 4 cos²θ + √3 = 2(√3 + 1) cos θ
Solution:
4 cos²θ – 2(√3 + 1) cos θ + √3 = 0
⇒ 4 cos²θ – 2√3 cos θ – 2 cos θ + √3 = 0
⇒ 2 cos θ (2 cos θ – √3) – 1(2 cos θ – √3) = 0
⇒ (2 cos θ – 1) (2 cos θ – √3) = 0
⇒ cos θ = \(\frac{1}{2}\) (or) cos θ = \(\frac{\sqrt{3}}{2}\)
If cos θ = \(\frac{1}{2}\) = cos \(\left(\frac{\pi}{3}\right)\)
∴ θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z is the general solution.
If cos θ = \(\frac{\sqrt{3}}{2}\) = cos \(\left(\frac{\pi}{6}\right)\)
∴ θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z is the general solution.

(vi) 1 + sin 2x – (sin 3x – cos 3x)²
Solution:
1 + sin 2x = (sin 3x – cos 3x)²
⇒ 1 + sin 2x = sin²3x + cos²3x – 2 sin 3x cos 3x
⇒ 1 + sin 2x = 1 – sin (2 × 3x)
⇒ sin 6x + sin 2x = 0
⇒ \(2 \sin \left(\frac{6 x+2 x}{2}\right) \cdot \cos \left(\frac{6 x-2 x}{2}\right)=0\)
⇒ sin (4x) . cos (2x) = 0
⇒ cos 2x = 0 (or) sin 4x = 0
If cos 2x = 0 = cos \(\frac{\pi}{2}\)
⇒ 2x = \(\frac{\pi}{2}\) is the principal solution and
2x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z is the general solution,
so that x = \(\frac{n \pi}{2}+\frac{\pi}{4}\), n ∈ Z
If sin 4x = 0 = sin(nπ), n ∈ Z
4x = nπ, n ∈ Z is the general solution.
So that 4x = nπ
⇒ x = \(\frac{n \pi}{4}\), n ∈ Z
∴ x = \(\frac{n \pi}{4}\); \(\frac{n \pi}{2}+\frac{\pi}{4}\), n ∈ Z is the general solution.

(vii) 2 sin²x + sin²2x = 2
Solution:
2 sin²x + sin²(2x) = 2
⇒ 2 sin2x + (2 sin x cos x)2 – 2 = 0
⇒ sin²x + 2 sin²x cos²x – 1 = 0
⇒ 2 sin²x cos²x – (1 – sin²x) = 0
⇒ 2 sin²x cos²x – cos²x = 0
⇒ cos²x (2 sin²x – 1) = 0
⇒ cos x = 0 (or) sin x = ±\(\frac{1}{\sqrt{2}}\)
If cos x = 0 = cos \(\frac{\pi}{2}\)
⇒ x = \(\frac{\pi}{2}\) is the principal solution
and cos x = 0 ⇔ \(\sin \left(x-\frac{\pi}{2}\right)=0\)
⇔ x – \(\frac{\pi}{2}\) = nπ, n ∈ Z
⇔ x = nπ + \(\frac{\pi}{2}\)
⇔ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
∴ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z is the general solution of cos x = 0
If sin x = ±\(\frac{1}{\sqrt{2}}\) = sin(±\(\frac{\pi}{4}\))
x = ±\(\frac{\pi}{4}\) are principal solutions
and x = nπ ± \(\frac{\pi}{4}\), n ∈ Z is the general solution
∴ The general solutions are \(\left[\left\{(2 n+1) \frac{\pi}{2}\right\},\left\{n \pi \pm \frac{\pi}{4}\right\} n \in Z\right]\)

Question 2.
Solve the following equations.
(i) √3 sin θ – cos θ = √2
Solution:

(ii) cot x + cosec x = √3
Solution:

(iii) sin x + √3 cos x = √2
Solution:

Question 3.
Solve the following equations:
(i) tan θ + sec θ = √3, 0 ≤ θ ≤ 2π
Solution:

(ii) cos 3x + cos 2x = sin \(\frac{3 x}{2}\) + sin \(\frac{x}{2}\), 0 ≤ x ≤ 2π
Solution:

(iii) cot²x – (√3 + 1) cot x + √3 = 0, 0 < x < \(\frac{\pi}{2}\)
Solution:
cot²x – (√3 + 1) cot x + √3 = 0
⇔ cot²x – √3 cot x – cot x + √3 = 0
⇔ cot x (cot x – √3) – 1(cot x – √3) = 0
⇔ cot x = √3 (or) cot x = 1
case (i): cot x = 1 ⇒ tan x = 1
∴ x = \(\left\{\frac{\pi}{4}\right\}\)
case (ii): cot x = √3 ⇒ tan x = \(\frac{1}{\sqrt{3}}\)
∴ x = \(\left\{\frac{\pi}{6}\right\}\)
∴ Solutions are \(\left\{\frac{\pi}{6}, \frac{\pi}{4}\right\}\)

(iv) sec x . cos 5x + 1 = 0; 0 < x < 2π
Solution:

III.

Question 1.
(i) Solve sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x
Solution:
(sin 3x + sin x) + sin 2x = (cos 3x + cos x) + cos 2x
⇔ 2 . sin(\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + sin 2x = 2 . cos (\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + cos 2x
⇔ 2 sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x
⇔ sin 2x (2 cos x + 1) = cos 2x (2 cos x + 1)
⇔ (2 cos x + 1) (sin 2x – cos 2x ) = 0
⇔ cos x = \(-\frac{1}{2}\) (or) sin 2x = cos 2x (i.e.,) tan (2x) = 1
Case (i):
cos x = \(-\frac{1}{2}\) = cos (\(\frac{2 \pi}{3}\))
Principal solution is x = \(\frac{2 \pi}{3}\)
and General solution is x = 2nπ ± \(\frac{2 \pi}{3}\), n ∈ z
Case (ii):
tan 2x = 1 = tan \(\frac{\pi}{4}\)
∴ Principal solution is 2x = \(\frac{\pi}{4}\) (i.e.,) x = \(\frac{\pi}{2}\)
General solution is 2x = nπ + \(\frac{\pi}{4}\), n ∈ z
⇒ x = \(\frac{\mathrm{n} \pi}{2}+\frac{\pi}{8}\), n ∈ Z
∴ General solution is \(\left.\left[\left\{2 n \pi \pm \frac{2 \pi}{3}\right\},\left\{\frac{n \pi}{2}+\frac{\pi}{8}\right\} / n \in Z\right\}\right]\)

(ii) If x + y = \(\frac{2 \pi}{3}\) and sin x + sin y = \(\frac{3}{2}\), find x and y.
Solution:

(iii) If sin 3x + sin x + 2 cos x = sin 2x + 2 cos²x, find the general solution.
Solution:
Given sin 3x + sin x + 2 cos x = sin 2x + 2 cos²x
⇒ 2 . sin (\(\frac{3 x+x}{2}\)) . cos (\(\frac{3 x-x}{2}\)) + 2 cos x = 2 sin x cos x + 2 cos²x
⇒ 2 . sin 2x . cos x + 2 cos x = 2 cos x (sin x + cos x)
⇒ 2 cos x (sin 2x + 1) = 2 cos x (sin x + cos x)
⇒ 2 cos x [sin 2x + 1 – sin x – cos x] = 0
⇒ cos x = 0 (or) sin 2x – sin x + 1 – cos x = 0

(iv) Solve cos 3x – cos 4x = cos 5x – cos 6x
Solution:
-2 sin 5x . sin x = -2 sin 4x . sin x
⇒ 2 sin x [sin 5x – sin 4x] = 0
⇒ 4 sin x . cos \(\frac{9 x}{2}\) . sin \(\frac{x}{2}\) = 0

Question 2.
Solve the following equations.
(i) cos 2θ + cos 8θ = cos 5θ
Solution:
cos 2θ + cos 8θ = cos 5θ
2 cos (\(\frac{2 \theta+8 \theta}{2}\)) cos (\(\frac{2 \theta-8 \theta}{2}\)) – cos 5θ = 0
2 cos 5θ . cos 3θ – cos 5θ = 0
cos 5θ (2 cos 3θ – 1) = 0
If cos 5θ = 0
Solution is 5θ = (2n + 1) \(\frac{\pi}{2}\)
θ = (2n + 1) \(\frac{\pi}{10}\), n ∈ z
If 2 cos 3θ – 1 = 0
cos 3θ = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
Solution is 3θ = 2nπ ± \(\frac{\pi}{3}\)
θ = \(\frac{2 n \pi}{3} \pm \frac{\pi}{9}\), n ∈ Z

(ii) cos θ – cos 7θ = sin 4θ
Solution:
-2 sin (\(\frac{\theta+7 \theta}{2}\)) sin (\(\frac{\theta-7 \theta}{2}\)) – sin 4θ = 0
2 sin 4θ sin 3θ – sin 4θ = 0
sin 4θ (2 sin 3θ – 1) = 0
If sin 4θ = 0
∴ Solution is 4θ = nπ
θ = \(\frac{n \pi}{4}\), n ∈ z
If 2 sin 3θ – 1 = 0
sin 3θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
Solution is 3θ = nπ + (-1)ⁿ \(\frac{\pi}{6}\)
θ = \(\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}\), n ∈ z

(iii) sin θ + sin 5θ = sin 3θ, 0 < θ < π
Solution:
sin θ + sin 5θ = sin 3θ
sin θ + sin 5θ – sin 3θ = 0
sin θ + 2 cos (\(\frac{5 \theta+3 \theta}{2}\)) sin (\(\frac{5 \theta-3 \theta}{2}\)) = 0
sin θ + 2 cos 4θ . sin θ = 0
sin θ (1 + 2 cos 4θ) = 0
sin θ = 0, cos 4θ = \(\frac{-1}{2}\)
If sin θ = 0, solution is θ = nπ, n ∈ Z
If cos 4θ = \(\frac{-1}{2}\) = cos(\(\frac{2 \pi}{3}\))
Solution is 4θ = 2nπ ± \(\frac{2 \pi}{3}\)
θ = \(\frac{2 n \pi}{4} \pm \frac{2 \pi}{12}\), n ∈ Z
θ = \(\frac{n \pi}{2} \pm \frac{\pi}{6}\), n ∈ Z
Since 0 < θ < π
Then π = \(\frac{\pi}{6}, \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{5 \pi}{6}\)

Question 3.
(i) If tan pθ = cot qθ and p ≠ -q show that the solutions are in A.P. with common difference \(\frac{\pi}{p+q}\)
Solution:
tan pθ = cot qθ = tan \(\frac{\pi}{2}\) – qθ
pθ = nπ + \(\frac{\pi}{2}\) – qθ
(p + q) θ = (2n + 1) \(\frac{\pi}{2}\)
θ = \(\frac{(2 n+1)}{p+q} \frac{\pi}{2}\), n is an integer
The Solutions \(\frac{\pi}{2(p+q)}, \frac{3 \pi}{2(p+q)}, \frac{5 \pi}{2(p+q)}\) + …………
∴ The solution form an Arithematical proportion With common difference \(\frac{2 \pi}{2(p+q)}=\frac{\pi}{p+q}\)

(ii) Show that the solutions of cos pθ = sin qθ form two series each of which is an A.P. Find also the common difference of each A.P. (p ≠ ±q).
Solution:
cos pθ = sin qθ
cos pθ – sin qθ = 0
cos pθ + \(\cos \left[\frac{\pi}{2}+9 \theta\right]\) = 0

(iii) Find the number of solutions of the equation tan x + sec x = 2 cos x; cos x ≠ 0, lying in the interival (0, π).
Solution:
tan x + sec x = 2 cos x
\(\frac{\sin x}{\cos x}+\frac{1}{\cos x}\) = 2 cos x
sin x + 1 = 2 cos²x
sin x + 1 = 2(1 – sin²x)
sin x + 1 = (2 – 2 sin²x)
2 sin²x + sin x – 1 = 0
2 sin²x + 2 sin x – sin x – 1 = 0
2 sin x (sin x + 1) – 1 (sin x + 1) = 0
(sin x + 1)(2 sin x – 1) = 0
sin x = -1 (or) sin x = \(\frac{1}{2}\)
If sin x = -1
x = \(\frac{-\pi}{2} \text { (or) } \frac{3 \pi}{2}\)
If sin x = \(\frac{1}{2}\)
x = \(\frac{\pi}{6} \text { (or) } \frac{5 \pi}{6}\)
In the interval (0, π)
Number of solutions = 2

(iv) Solve sin 3α = 4 sin α sin(x + α) sin(x – α) where α ≠ nπ, n ∈ Z
Solution:
3 sin α – 4 sin³α = 4 sin α (sin²x – sin²α)
Dividing with sin α
3 – 4 sin²α = 4 (sin²x – sin²α)
3 – 4 sin²α = 4 sin²x – 4 sin²α
4 sin²x = 3
2 sin²x = \(\frac{3}{2}\)
1 – cos 2x = \(\frac{3}{2}\)
cos 2x = \(-\frac{1}{2}=\cos \frac{2 \pi}{3}\)
2x = 2nπ ± \(\frac{2 \pi}{3}\), ∀ n ∈ Z
x = nπ ± \(\frac{\pi}{3}\), n ∈ Z

Question 4.
(i) If tan(π cos θ) = cot(π sin θ), then prove that \(\cos \left(\theta-\frac{\pi}{4}\right)=\pm \frac{1}{2 \sqrt{2}}\)
Solution:

(ii) Find the range of θ if cos θ + sin θ is positive.
Solution:

Question 5.
If α, β are the solutions of the equation a cos θ + b sin θ = c, where a, b, c ∈ R and if a² + b² > 0, cos α ≠ cos β and sin α ≠ sin β, then show that
(i) sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
(ii) cos α + cos β = \(\frac{2 a c}{a^{2}+b^{2}}\)
(iii) cos α . cos β = \(\frac{c^{2}-b^{2}}{a^{2}+b^{2}}\)
(iv) sin α . sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)
Solution:
a cos θ = b sin θ = c
First write this as a quadratic equation in sin θ
⇔ a cos θ = c – b sin θ
By squaring on both sides, we get
a² cos²θ = (c – b sin θ)²
⇔ a² (1 – sin²θ) = c² + b² sin²θ – 2 bc sin θ
⇔ (a² + b²) sin²θ – 2bc sin θ + (c² – a²) = 0
It is a quadratic equation in sin θ,
It has sin α and sin β as roots since α and β are solutions for the given equation
(i) Sum of the roots = sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
Again a cos θ + b sin θ = c
Write this as a quadratic equation in cos θ
⇔ b sin θ = c – a cos θ
By squaring on both sides
⇔ b² sin²θ = (c – a cos θ)²
⇔ b²(1 – cos²θ) = c² + a² cos²θ – 2 ca cos θ
⇔ (a² + b²) cos²θ – 2 ca cos θ + (c² – b²) = 0
It is a quadratic equation in cos θ. It has cos α, cos β be its roots.
(ii) Sum of the roots = cos α + cos β = \(\frac{2 c a}{a^{2}+b^{2}}\)
(iii) Product of the roots = cos α . cos β = \(\frac{c^{2}-b^{2}}{a^{2}+b^{2}}\)
(iv) Product of the roots = sin α . sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)

Question 6.
(i) Find the common roots of the equations cos 2x + sin 2x = cot x and 2 cos²x + cos²2x = 1.
Solution:
Let tan x = A
Given that cos 2x + sin 2x = cot x
⇔ \(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}+\frac{2 \tan x}{1+\tan ^{2} x}=\frac{1}{\tan x}\)
⇔ \(\frac{1-A^{2}}{1+A^{2}}+\frac{2 A}{1+A^{2}}=\frac{1}{A}\)
⇔ (1 – A² + 2A) A = (1 + A2)
⇔ A – A³ + 2A² = 1 + A²
⇔ A³ – A² – A + 1 = 0
⇔ A = 1
A = 1 satisfy this equation

∴ A³ – A² – A + 1 = 0
⇔ (A – 1) (A² – 1) = 0
⇔ (A – 1) (A – 1) (A + 1) = 0
⇔ A = 1, A = -1
∴ tan x = ±1
⇒ x = (2n + 1) \(\frac{\pi}{4}\), n ∈ z
Given 2 cos²x + cos² 2x = 1
⇔ (2 cos²x – 1) + cos²(2x) = 0
⇔ cos 2x + cos²(2x) = 0
⇔ cos 2x (1 + cos 2x) = 0
⇔ cos 2x = 0 (or) cos 2x = -1
Case (i): cos 2x = 0
⇔ 2x = (2n + 1) \(\frac{\pi}{2}\)
∴ x = (2n + 1) \(\frac{\pi}{4}\), n ∈ z
∴ (2n + 1) \(\frac{\pi}{4}\), n ∈ z is the common root for the given two trigonometric equations.

(ii) Solve the equation \(\sqrt{6-\cos x+7 \sin ^{2} x}+\cos x=0\)
Solution:
\(\sqrt{6-\cos x+7 \sin ^{2} x}+\cos x=0\)
6 – cos x + 7 sin²x ≥ 0
⇒ 7(1 – cos²x) – cos x + 6 ≥ 0
⇒ 7 – 7 cos²x – cos x + 6 ≥ 0
⇒ 7 cos²x + cos x – 13 ≤ 0
Consider 7 cos²x + cos x – 13 = 0

cos x values do not lie in [-1, 1]
Hence the given equation has no solution.

(iii) If |tan x| = tan x + \(\frac{1}{\cos x}\) and x ∈ [0, 2π], find the value of x.
Solution:
Case (i):
|tan x| = tan x, if x lies either in I (or) in III quadrant
Then |tan x| = tan x + \(\frac{1}{\cos x}\)
⇒ tan x = tan x + sin x
⇒ sec x = 0 which is impossible, since sec x ∉ (-1, 1)
Case (ii):
|tan x| = -tan x, if x lies in II & IV quadrants
Then |tan x| = tan x + \(\frac{1}{\cos x}\)
⇒ -tan x = tan x + sec x
⇒ -2 tan x = sec x
⇒ \(-2 \frac{\sin x}{\cos x}-\frac{1}{\cos x}\) = 0
⇒ -2 sin x – 1 = 0
⇒ sin x = \(\frac{-1}{2}=\sin \left(\frac{-\pi}{6}\right)=\sin \left(2 \pi-\frac{\pi}{6}\right)\)
∴ x = \(\frac{11 \pi}{6}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(f)

Question 1.
If A, B, C are angles in a triangle, then prove that
(i) sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C
Solution:
∵ A, B, C are angles in a triangle
⇒ A + B + C = 180° ……….(1)
LHS = sin 2A – sin 2B + sin 2C
= sin 2A + sin 2C – sin 2B
= 2 sin (\(\frac{2 A+2 C}{2}\)) . cos(\(\frac{2 A-2 C}{2}\)) – sin 2B
= 2 sin (A + C) cos (A – C) – sin B
= 2 sin (180° – B) cos (A – C) – 2 sin B cos B
= 2 sin B cos (A – C) – 2 sin B cos B
= 2 sin B [cos (A – C) – cos B]
= 2 sin B [cos (A – C) – cos (180° – (A + C)]
= 2 sin B [cos (A – C) + cos (A + C)]
= 2 sin B (2 cos A cos C)
= 4 cos A sin B cos C
∴ sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C

(ii) cos 2A – cos 2B + cos 2C = 1 – 4 sin A cos B Sin C
Solution:
L.H.S. = -(cos 2B – cos 2A) + cos 2C
= -2 sin (A + B) sin (A – B) + cos 2C
= -2 sin (180° – C) sin (A – B) + cos 2C
= -2 sin C sin (A – B) + 1 – 2 sin2C
= 1 – 2 sin C (sin (A – B) + sin C)
= 1 – 2 sin C sin (A – B) + sin (180° – \(\overline{\mathrm{A}+\mathrm{B}}\))
= 1 – 2 sin C (sin (A – B) + sin (A + B))
= 1 – 2 sin C (2 sin A cos B)
= 1 – 4 sin A cos B sin C
= R.H.S.

Question 2.
If A, B, C are angles in a triangle, then prove that
(i) sin A + sin B – sin C = 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) cos \(\frac{C}{2}\)
Solution:
L.H.S. = (sin A + sin B) – sin C
= 2 sin (\(\frac{A+B}{2}\)) cos (\(\frac{A-B}{2}\)) – sin C

(ii) cos A + cos B – cos C = -1 + 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\)
Solution:
A, B, C are angles in a triangle
A + B + C = 180° ………(1)
LHS = cos A + cos B – cos C

Question 3.
If A, B, C are angles in a triangle, then prove that
(i) sin²A + sin²B – sin²C = 2 sin A sin B cos C
Solution:
Given A + B + C = 180°
L.H.S. = sin²A + [sin²B – sin²C]
= sin²A + sin (B + C) sin (B – C)
= sin²A + sin (180° – A) . sin (B – C)
= sin²A + sin A . sin (B – C)
= sin A (sin A + sin (B – C))
= sin A [sin (180° – \(\overline{\mathrm{B}+\mathrm{C}}\)) + sin (B – C)]
= sin A [sin (B + C) + sin (B – C)]
= sin A [2 sin B cos C]
= 2 sin A sin B cos C
= R.H.S

(ii) cos²A + cos²B – cos²C = 1 – 2 sin A sin B cos C
Solution:
A, B, C are angles in a triangle
⇒ A + B + C = 180° ……..(1)
L.H.S = cos²A + cos²B – cos²C
= cos²A + cos²B – cos²C

= 1 + cos (A + B) cos (A – B) – cos²C
= 1 + cos (180° – C) cos (A – B) – cos²C [By (1)]
= 1 – cos C cos (A – B) – cos²C
= 1 – cos C [cos (A – B) + cos C]
= 1 – cos C [cos (A – B) + cos(180° – \(\overline{A+B}\)] [By eq. (1)]
= 1 – cos C [cos (A – B) – cos (A + B)]
= 1 – cos C [2 sin A sin B]
= 1 – 2 sin A sin B cos C
∴ cos²A + cos²B – cos²C = 1 – 2 sin A sin B cos C

Question 4.
If A + B + C = π, then prove that
(i) \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}\) = \(2\left[1+\sin \frac{A}{2}+\sin \frac{B}{2} \sin \frac{C}{2}\right]\)
Solution:

(ii) \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}-\cos ^{2} \frac{C}{2}=2 \cos \frac{A}{2}\) \(\cos \frac{B}{2} \sin \frac{C}{2}\)
Solution:

Question 5.
In triangle ABC, prove that
(i) \(\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2}\) = \(4 \cos \frac{\pi-A}{4} \cos \frac{\pi-B}{4} \cos \frac{\pi-C}{4}\)
Solution:

(ii) \(\cos \frac{A}{2}+\cos \frac{B}{2}-\cos \frac{C}{2}\) = \(4 \cos \frac{\pi+A}{4} \cdot \cos \frac{\pi+B}{4} \cos \frac{\pi-C}{4}\)
Solution:

(iii) \(\sin \frac{A}{2}+\sin \frac{B}{2}-\sin \frac{C}{2}\) = \(-1+4 \cos \frac{\pi-A}{4} \cos \frac{\pi-B}{4} \sin \frac{\pi-C}{4}\)
Solution:

Question 6.
If A + B + C = π/2, then prove that cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Solution:
A + B + C = π/2 ………(1)
LHS = cos 2A + cos 2B + cos 2C
= 2 cos (\(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\)) cos (\(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\)) + cos 2C
= 2 cos (A + B) . cos (A – B) + cos 2C
= 2 cos (90° – C) cos (A – B) + cos 2C
= 2 sin C cos (A – B) + (1 – 2 sin²C)
= 1 + 2 sin C [cos (A – B) – sin C]
= 1 + 2 sin C [cos (A – B)- sin (90° – \(\overline{A+B}\))]
= 1 + 2 sin C [cos (A – B) – cos (A +B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C
= RHS
∴ cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C

Question 7.
If A + B + C = 3π/2, then prove that
(i) cos²A + cos²B – cos²C = -2 cos A cos B sin C
Solution:
A + B + C = 3π/2 ……..(1)
L.H.S. = cos²A + cos²B – cos²C
= cos²A + (1 – sin²B) – cos²C
= (cos²A – sin²B) + (1 – cos²C)
= cos (A + B) cos (A – B) + sin²C
= cos (270° – C) cos(A – B) + sin²C
= -sin C cos (A – B) + sin²C
= sin C [sin C – cos (A – B)]
= sin C [sin (270°- \(\overline{A+B}\)) – cos (A – B)]
= sin C [-cos (A + B) – cos (A – B)]
= -sin C [cos (A + B) + cos (A – B)]
= -sin C [2 cos A cos B]
= -2 cos A cos B sin C
= RHS
∴ cos²A + cos²B – cos²C = -2 cos A cos B sin C

(ii) sin 2A + sin 2B – sin 2C = -4 sin A sin B cos C
Solution:
Here A + B + C = 270° ………(1)
LHS = sin 2A + sin 2B – sin 2C
= 2 sin (\(\frac{2 A+2 B}{2}\)) cos (\(\frac{2 A-2 B}{2}\)) – sin 2C
= 2 sin (A + B) . cos (A – B) – 2 sin C cos 2C
= 2 sin (270° – C) cos (A – B) – 2 sin C cos C
= -2 cos C cos (A – B) – 2 sin C cos C
= -2 cos C [cos (A – B) + sin C]
= -2 cos C [cos (A – B) + sin (270° – \(\overline{A+B}\))]
= -2 cos C [cos (A – B) – cos (A + B)]
= -2 cos C [2 sin A sin B]
= -4 sin A sin B cos C
= RHS
∴ sin 2A + sin 2B – sin 2C = -4 sin A sin B cos C

Question 8.
If A + B + C = 0°, then prove that
(i) sin 2A + sin 2B + sin 2C = -4 sin A sin B sin C
Solution:
Here A + B + C = 0 ………(1)
LHS = sin 2A + sin 2B + sin 2C
= 2 sin (\(\frac{2 A+2 B}{2}\)) cos (\(\frac{2 A-2 B}{2}\)) + sin 2C
= 2 sin (A + B) . cos (A – B) + 2 sin C cos 2C
= 2 sin (-C) cos (A – B) + 2 sin C cos C
= -2 sin C cos (A – B) + 2 sin C cos C
= -2 sin C [cos (A – B) – cos C]
= -2 sin C [cos (A – B) – cos (-A – B)]
= -2 sin C [cos (A – B) – cos (A + B)]
= -2 sin C [2 sin A sin B]
= -4 sin A sin B sin C
= RHS
∴ sin 2A + sin 2B + sin 2C = -4 sin A sin B sin C

(ii) sin A + sin B – sin C = – 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\)
Solution:

Question 9.
If A + B + C + D = 2π then prove that
(i) sin A – sin B + sin C – sin D = \(-4 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+D}{2}\right)\)
Solution:

(ii) cos 2A + cos 2B + cos 2C + cos 2D = 4 cos (A + B) cos (A + C) cos (A + D)
Solution:

Question 10.
If A + B + C = 2S, then prove that
(i) sin (S – A) + sin (S – B) + sin C = \(4 \cos \left(\frac{S-A}{2}\right) \cos \left(\frac{S-B}{2}\right) \sin \frac{C}{2}\)
Solution:

(ii) cos (S – A) + cos (S – B) + cos C = \(-1+4 \cos \left(\frac{S-A}{2}\right) \cos \left(\frac{S-B}{2}\right) \cdot \cos \frac{C}{2}\)
Solution: