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

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

I. Convert the following into the simplest form.

Question 1.
(i) tan(θ – 14π)
Solution:
= tan(14π – θ)
= tan(2 . (7π) – θ)
= tan θ

(ii) \(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
Solution:
\(\cot \left(\frac{21 \pi}{2}-\theta\right)\)
= \(\cot \left(10 \pi+\left(\frac{\pi}{2}-\theta\right)\right)\)
= \(\cot \left(\frac{\pi}{2}-\theta\right)\)
= tan θ

(iii) cosec(5π + θ)
Solution:
cosec(5π + θ) = cosec(2π + (3π + θ))
= cosec(3π + θ)
= cosec(2π + (π + θ))
= cosec(π + θ)
= -cosec θ

(iv) sec(4π – θ?)
Solution:
sec(4π – θ)
= sec(2π + (2π – θ))
= sec(2π – θ)
= sec θ

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 2.
Find the value of each of the following.
(i) sin(-405°)
Solution:
sin(-405°) = -sin(360° + 45)
= -sin 45°
= \(-\frac{1}{\sqrt{2}}\)

(ii) \(\cos \left(-\frac{7 \pi}{2}\right)\)
Solution:
\(\cos \left(-\frac{7 \pi}{2}\right)\)
= -cos 630°
= -cos(360° + 270°)
= -cos 270°
= -cos(180° + 90°)
= -cos 90°
= 0

(iii) sec(2100°)
Solution:
sec(2100°) = sec (5 × 360° + 300°)
= sec 300°
= sec(360 – 60°)
= sec 60°
= 2

(iv) cot(-315°)
Solution:
cot(-315°) = -cot 315°
= -cot(360° – 45°)
= -cot 45°
= 1

Question 3.
Evaluate.
(i) cos2 45° + cos2 135° + cos2 225° + cos2 315°
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) I Q3(i)

(ii) \(\sin ^{2} \frac{2 \pi}{3}+\cos ^{2} \frac{5 \pi}{6}-\tan ^{2} \frac{3 \pi}{4}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) I Q3(ii)

(iii) cos 225° – sin 225° + tan 495° – cot 495°
Solution:
cos (180° + 45°) – sin(180° + 45°) + tan(360° + 135°) – cot(360° + 135°)
= -cos 45° + sin 45° – tan 135° + cot 135°
= \(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1-1\)
= 0

(iv) (cos θ – sin θ) if (a) θ = \(\frac{7 \pi}{4}\) (b) θ = \(\frac{11 \pi}{3}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) I Q3(iv)

Question 4.
(i) If sin θ = \(\frac{-1}{3}\) and θ does not lie in the third quadrant, find the values of (a) cos θ (b) cot θ
Solution:
∵ sin θ = \(\frac{-1}{3}\) and sin θ is negative and θ does not lie in the IIIrd quadrant.
⇒ θ lies in IV quadrant.
∴ In the IV quadrant, cos θ is positive and cot θ is negative.
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) I Q4(i)

(ii) If cos θ = t (0 < t < 1) and θ does not lie in the first quadrant, find the values of (a) sin θ (b) tan θ.
Solution:
cos θ = t, (0 < t < 1)
⇒ cos θ is positive and θ does not lie in the first quadrant.
⇒ θ lies in IV quadrant.
(a) sin θ = \(-\sqrt{1-\cos ^{2} \theta}=-\sqrt{1-t^{2}}\)
(b) tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{-\sqrt{1-t^{2}}}{t}\)

(iii) Find the value of sin 330°. cos 120° + cos 210°. sin 300°.
Solution:
sin 330°. cos 120° + cos 210°. sin 300°
= sin(360° – 30°) . cos(180° – 60°) + cos(180° + 30°) . sin(360° – 60°)
= (-sin 30°) (-cos 60°) + (-cos 30°) (-sin 60°)
= sin 30° cos 60° + cos 30° sin 60°
= sin(30° + 60°) [∵ sin A cos B + cos A sin B = sin(A + B)]
= sin (90°)
= 1

(iv) If cosec θ + cot θ = \(\frac{1}{3}\), find cos θ and determine the quadrant in which θ lies.
Solution:
cosec θ + cot θ = \(\frac{1}{3}\)
⇒ cosec θ – cot θ = 3 (∵ cosec2θ – cot2θ = 1)
∴ 2 cosec θ = \(\frac{1}{3}\) + 3 = \(\frac{10}{3}\)
∴ cosec θ = \(\frac{5}{3}\) and sin θ = \(\frac{3}{5}\)
2 cot θ = \(\frac{1}{3}\) – 3 = \(\frac{-8}{3}\)
∴ cot θ = \(\frac{-4}{3}\), tan θ = \(\frac{-3}{4}\)
cos θ = (cot θ) . sin θ = \(\left(\frac{-4}{3}\right)\left(\frac{3}{5}\right)=\frac{-4}{5}\)
∵ sin θ is +ve and cos θ is -ve
⇒ θ lies in II quadrant.

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 5.
(i) If sin α + cosec α = 2. Find the value of sinnα + cosecnα, n ∈ z.
Solution:
Given sin α + cosec α = 2
Squaring on both sides
sin2α + cosec2α + 2 = 4
sin2α + cosec2α = 2
sin α + cosec α = 2
Cubing on both sides
sin3α + cosec3α + 3 sin α . cosec α (sin α + cosec α) = 8
sin3α + cosec3α + 3(2) = 8
sin3α + cosec3α = 8 – 6
sin3α + cosec3α = 2
similarly sinnα + cosecnα = -2

(ii) If sec θ + tan θ = 5. Find the quadrant in which θ lies and find the value of sin θ.
Solution:
sec θ + tan θ = 5
⇒ sec θ – tan θ = \(\frac{1}{5}\) (∵ sec2θ – tan2θ = 1)
2 sec θ = 5 + \(\frac{1}{5}\) = \(\frac{26}{5}\)
sec θ = \(\frac{26}{10}=\frac{13}{5}\)
Aquir 2 tan θ = 5 – \(\frac{1}{5}\) = \(\frac{24}{5}\)
tan θ = \(\frac{24}{10}=\frac{12}{5}\)
Now sin θ = \(\frac{\tan \theta}{\sec \theta}=\frac{\frac{12}{5}}{\frac{13}{5}}=\frac{12}{13}\)
tan θ is +ve, sec θ is +ve
⇒ θ lies in the first quadrant.

II.

Question 1.
Prove that
(i) \(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\) = cos A
Solution:
\(\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left[\frac{3 \pi}{2}+A\right] \sin (2 \pi-A)}\)
= \(\frac{-\cos A(-\tan A) \cos A}{\tan A(-\cot A)(-\sin A)}\)
= cos A

(ii) \(\frac{\sin (3 \pi-A) \cos \left(A-\frac{\pi}{2}\right) \tan \left(\frac{3 \pi}{2}-A\right)}{{cosec}\left(\frac{13 \pi}{2}+A\right) \sec (3 \pi+A) \cot \left(A-\frac{\pi}{2}\right)}\) = cos4A
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q1(ii)

(iii) sin 780° sin 480° + cos 240°. cos 300° = \(\frac{1}{2}\)
Solution:
sin(2(360°) + 60°) . sin(360° + 120°) + cos(270° – 30°) . cos(360° – 60°)
= sin 60° . sin 120° – sin 30° . cos 60°
= \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{1}{2}\)
= \(\frac{3}{4}-\frac{1}{4}\)
= \(\frac{1}{2}\)

(iv) \(\frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135^{\circ}+3 \sin 210^{\circ}}\) = -2
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q1(iv)

(v) cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\)) = 1
Solution:
L.H.S. = cot(\(\frac{\pi}{20}\)) . cot(\(\frac{3\pi}{20}\)) . cot(\(\frac{5\pi}{20}\)) . cot(\(\frac{7\pi}{20}\)) . cot(\(\frac{9\pi}{20}\))
= cot (9°) cot (27°) cot (45°) cot (63°) cot (81°)
= cot (9°) cot (27°) (1) cot (90° – 27°) cot (90° – 9°)
= cot (9°) cot (27°) tan 27° tan 9°
= (tan 9° cot 9°) (tan 27° cot 27°)
= (1) (1)
= 1

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 2.
Simplify.
(i) \(\frac{\sin \left(-\frac{11 \pi}{3}\right) \tan \left(\frac{35 \pi}{6}\right) \sec \left(-\frac{7 \pi}{3}\right)}{\cot \left(\frac{5 \pi}{4}\right) {cosec}\left(\frac{7 \pi}{4}\right) \cos \left(\frac{17 \pi}{6}\right)}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q2(i)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q2(i).1

(ii) If tan 20° = p, prove that \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}=\frac{1-p^{2}}{1+p^{2}}\)
Solution:
Given tan 20° = p
L.H.S. = \(\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}\)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q2(ii)

(iii) If α, β are complementary angles such that b sin α = a, then find the value of (sin α cos β – cos α sin β).
Solution:
∵ α, β are complementary angles
⇒ α + β = 90°
⇒ β = 90° – α
Now sin α cos β – cos α sin β
= sin (α – β)
= sin [(α – (90° – α)]
= sin [2α – 90°]
= -sin (90° – 2α)
= -cos 2α
= -(1 – 2 sin2α) (∵ cos 2α = 1 – 2 sin2α)
= \(-1+2\left(\frac{a}{b}\right)^{2}\) (∵ sin α = \(\frac{a}{b}\))
= \(\frac{-b^{2}+2 a^{2}}{b^{2}}\)
= \(\frac{2 a^{2}-b^{2}}{b^{2}}\)

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 3.
(i) If cos A = cos B = \(-\frac{1}{2}\), A does not lie in the second quadrant and B does not lie in the third quadrant, then find the value of \(\frac{4 \sin B-3 \tan A}{\tan B+\sin A}\)
Solution:
∵ cos A = \(-\frac{1}{2}\) and A does not lie in second quadrant.
⇒ A lies in the third quadrant, (∵ cos A is -ve)
and cos B = \(-\frac{1}{2}\) and B does not lie in third quadrant.
⇒ B lies in second quadrant.
∵ cos A = \(-\frac{1}{2}\) and A lies in third quadrant.
⇒ A = 240°
∵ cos B = \(-\frac{1}{2}\) and B lies in second quadrant.
⇒ B = 120°
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q3(i)

(ii) If 8 tan A = -15 and 25 sin B = -7 and neither A nor B is in the fourth quadrant, then show that sin A cos B + cos A sin B = \(\frac{-304}{425}\)
Solution:
8 tan A = -15 ⇒ tan A = \(\frac{-15}{8}\)
25 sin B = -7 ⇒ sin B = \(-\frac{7}{25}\)
Given neither A nor B is in the fourth quadrant.
Clearly, A is the second quadrant B is the third quadrant
sin A cos B + cos A sin B = \(\left(\frac{15}{17}\right)\left(\frac{-24}{25}\right)+\left(\frac{-8}{17}\right)\left(\frac{-7}{25}\right)\)
= \(\frac{-360}{425}+\frac{56}{425}\)
= \(\frac{-304}{425}\)

(iii) If A, B, C, D are angles of a cyclic quadrilateral, then prove that
(a) sin A – sin C = sin D – sin B
(b) cos A + cos B + cos C + cos D = 0
Solution:
∵ A, B, C, D are angles of a cyclic quadrilateral.
⇒ A + C = 180° and B + D = 180° ……..(1)
C = 180° – A and D = 180° – B
(a) L.H.S. = sin A – sin C
= sin A – sin(180° – A)
= sin A – sin A
= 0
R.H.S. = sin D – sin B
= sin(180°- B) – sin B
= sin B – sin B
= 0
∴ L.H.S. = R.H.S.
i.e., sin A – sin C = sin D – sin B
(b) L.H.S. = cos A + cos B + cos C + cos D
= cos A + cos B + cos(180° – A) + cos(180° – B)
= cos A + cos B – cos A – cos B
= 0
∴ cos A + cos B + cos C + cos D = 0

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 4.
(i) If a cos θ – b sin θ = c, then show that a sin θ + b sin θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\).
Solution:
a cos θ – b sin θ = c
let a sin θ + b cos θ = x
by squaring and adding
(a cos θ – b sin θ)2 + (a sin θ + b cos θ)2 = c2 + x2
a2 cos2θ + b2 sin2θ – 2ab sin θ cos θ + a2 sin2θ + b2 cos2θ + 2ab sin θ cos θ = c2 + x2
a2 + b2 = c2 + x2
a2 + b2 – c2 = x2
x = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)
∴ a sin θ + b cos θ = \(\pm \sqrt{a^{2}+b^{2}-c^{2}}\)

(ii) If 3 sin A + 5 cos A = 5, then show that 5 sin A – 3 cos A = ±3.
Solution:
3 sin A + 5 cos A = 5
Let 5 sec A – 3 cos A = x
by squaring and adding
(3 sin A + 5 cos A)2 + (5 sin A – 3 cos A)2 = 52 + x2
9 sin2 A + 25 cos2 A + 30 sin A cos A + 25 sin2 A + 9 cos2 A – 30 sin A cos A = 25
9 + 25 = 25 + x2
x2 = 9
x = ±3
∴ 5 sin A – 3 cos A = ±3

(iii) If tan2θ = (1 – e2), show that sec θ + tan3θ . cosec θ = (2 – e2)3/2.
Solution:
tan2θ = 1 – e2
sec2θ = 1 + tan2θ = 2 – e2
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) II Q4(iii)

III. Prove the following.

Question 1.
(i) \(\frac{(\tan \theta+\sec \theta-1)}{(\tan \theta-\sec \theta+1)}=\frac{1+\sin \theta}{\cos \theta}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q1(i)

(ii) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2.
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q1(ii)

(iii) 3(sin θ – cos θ)4 + 6(sin θ + cos θ)2 + 4(sin6θ + cos6θ) = 13.
Solution:
(sin θ – cos θ)2 = sin2θ + cos2θ – 2 sin θ . cos θ
= 1 – 2 sin θ cos θ
(sin θ – cos θ)4 = (1 – 2 sin θ cos θ)2
= 1 + 4 sin2θ cos2θ – 4 sin θ cos θ ……(1)
(sin θ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ
= 1 + 2 sin θ cos θ …….(2)
sin6θ + cos6θ = (sin2θ +cos2θ)3 – 3 sin2θ cos2θ (sin2θ + cos2θ)
= 1 – 3 sin2θ cos2θ ……..(3)
L.H.S. = 3(1 + 4 sin2θ cos2θ – 4 sin θ cos θ) + 6(1 + 2 sin θ cos θ) + 4(1 – 3 sin2θ cos2θ)
= 3 + 12 sin2θ cos2θ – 12 sin θ cos θ + 6 + 12 sin θ cos θ + 4 – 12 sin2θ cos2θ
= 3 + 6 + 4
= 13
= R.H.S.

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 2.
(i) Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 – (tan2θ + cot2θ) = 7.
Solution:
L.H.S. = (sin θ + cosec θ)2 + (cos θ + sec θ)2 – (tan2θ + cot2θ)
= (sin2θ + cosec2θ + 2 sin θ cosec θ) + (cos2θ + sec2θ + 2 cos θ sec θ) – (tan2θ + cot2θ)
= (sin2θ + cos2θ) + (1 + cot2θ) + (1 + tan2θ) + 4 – tan2θ – cot2θ
= 1 + 1 + 1 + 4
= 7

(ii) cos4α + 2 cos2α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\) = (1 – sin4α)
Solution:
L.H.S. = cos4α + 2 cos2α \(\left(1-\frac{1}{\sec ^{2} \alpha}\right)\)
= cos4α + 2 cos2α (1 – cos2α)
= cos2α [cos2α + 2 sin2α]
= (1 – sin2α) [cos2α + sin2α + sin2α]
= (1 – sin2α) (1 + sin2α)
= 1 – sin4α

(iii) \(\frac{(1+\sin \theta-\cos \theta)^{2}}{(1+\sin \theta+\cos \theta)^{2}}=\frac{1-\cos \theta}{1+\cos \theta}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q2(iii)

(iv) If \(\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\) = x, then find the value of \(\frac{(1-\cos \theta+\sin \theta)}{(1+\sin \theta)}\)
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q2(iv)

Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a)

Question 3.
Eliminate θ from the following.
(i) x = a cos3θ; y = b sin3θ
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q3(i)

(ii) x = a cos4θ; y = b sin4θ
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q3(ii)

(iii) x = a(sec θ + tan θ); y = b(sec θ – tan θ)
Solution:
\(\frac{x}{a}\) = sec θ + tan θ; \(\frac{y}{b}\) = (sec θ – tan θ)
\(\frac{x}{a} \times \frac{y}{b}\) = (sec θ + tan θ) (sec θ – tan θ)
\(\frac{x y}{a b}\) = sec2θ – tan2θ
\(\frac{x y}{a b}\) = 1
xy = ab

(iv) x = cot θ + tan θ; y = sec θ – cos θ
Solution:
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q3(iv)
Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(a) III Q3(iv).1