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

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:

(ii) $$\sin ^{2} \frac{2 \pi}{3}+\cos ^{2} \frac{5 \pi}{6}-\tan ^{2} \frac{3 \pi}{4}$$
Solution:

(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:

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.

(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.

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:

(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:

(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

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:

(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. = $$\frac{\tan 610^{\circ}+\tan 700^{\circ}}{\tan 560^{\circ}-\tan 470^{\circ}}$$

(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 = $$\frac{a}{b}$$)
= $$\frac{-b^{2}+2 a^{2}}{b^{2}}$$
= $$\frac{2 a^{2}-b^{2}}{b^{2}}$$

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°

(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.
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

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
(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
(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

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:

(ii) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2.
Solution:

(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.

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:

(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:

Question 3.
Eliminate θ from the following.
(i) x = a cos3θ; y = b sin3θ
Solution:

(ii) x = a cos4θ; y = b sin4θ
Solution:

(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: