GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

   

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Prove that:
1. sin2 \(\frac{π}{6}\) + cos2 \(\frac{π}{3}\) – tan2 \(\frac{π}{4}\) = – \(\frac{1}{2}\)
2. 2 sin2 \(\frac{π}{6}\) + cosec2 \(\frac{7π}{6}\) cos2 \(\frac{π}{3}\) = \(\frac{3}{2}\)
3. cot2 \(\frac{π}{6}\) + cosec \(\frac{5π}{6}\) + 3 tan2 \(\frac{π}{6}\) = 6
4. 2 sin2 \(\frac{3π}{4}\) + 2 cos2 \(\frac{π}{4}\) + 2 sec2 \(\frac{π}{3}\) = 10
Solutions to questions 1 – 4:
1. L.H.S. = sin2 \(\frac{π}{6}\) + cos2 \(\frac{π}{3}\) – tan2 \(\frac{π}{4}\)
= (sin \(\frac{π}{6}\))2 + (cos \(\frac{π}{3}\))2 – (tan \(\frac{π}{4}\))2
= (\(\frac{1}{2}\))2 + (\(\frac{1}{2}\))2 – 12
[∵ sin \(\frac{π}{6}\) = \(\frac{1}{2}\), cos \(\frac{π}{3}\) = \(\frac{1}{2}\), tan \(\frac{π}{4}\) = 1]
= \(\frac{1}{4}\) + \(\frac{1}{4}\) – 1 = \(\frac{1}{2}\) – 1
= \(\frac{- 1}{2}\) = R.H.S.

2. L.H.S. = 2 sin2 \(\frac{π}{6}\) + cosec2 \(\frac{7π}{6}\) cos2 \(\frac{π}{3}\)
= 2(sin \(\frac{π}{6}\))2 + (cosec \(\frac{7π}{6}\))2 (cos \(\frac{π}{3}\))2
= 2(\(\frac{1}{2}\))2 + [cosec (π + \(\frac{π}{6}\))]2 (\(\frac{1}{2}\))2
= 2 × \(\frac{1}{4}\) + (- cosec \(\frac{π}{6}\))2 ((\(\frac{1}{4}\)) [∵ cosec (π + θ) = – cosec θ]
= \(\frac{1}{2}\) + (2)2 \(\frac{1}{4}\) = \(\frac{1}{2}\) + 1 = \(\frac{3}{2}\) = R.H.S.

3. L.H.S. = cot2 \(\frac{π}{6}\) + cosec \(\frac{5π}{6}\) + 3 tan2 \(\frac{π}{6}\)
= (cot \(\frac{π}{6}\))2 + cosec (π – \(\frac{π}{6}\)) + 3(tan \(\frac{π}{6}\))2
= (\(\sqrt{3}\))2 + cosec \(\frac{π}{6}\) + 3(\(\frac{1}{\sqrt{3}}\))2
= 3 + 2 + 3 × \(\frac{1}{3}\) = 3 + 2 + 1 = 6 = R.H.S.

4. L.H.S. = 2 sin2 \(\frac{3π}{4}\) + 2 cos2 \(\frac{π}{4}\) + 2 sec2 \(\frac{π}{3}\)
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 1
= 1 + 1 + 8 = 10 = R.H.S.

5. Find the value of:
(i) sin 75°
(ii) tan 15°
Solution:
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30° [sin (A + B) = sin A cos B + cos A sin B]
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 2

(ii) tan 15° = tan (45° – 30°)
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 3

GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Prove the following:
6. cos (\(\frac{π}{4}\) – x) cos (\(\frac{π}{4}\) – y) – sin (\(\frac{π}{4}\) – x) sin (\(\frac{π}{4}\) – y) = sin (x + y)
7. \(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\frac{(1+\tan x)^{2}}{(1-\tan x)^{2}}\)
8. \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\) = cot2 x
9. cos (\(\frac{3π}{2}\) + x)cos (2π + x) [cos ((\(\frac{3π}{2}\) – x) + cot (2π + x)] = 1
Solutions to questions 6 – 9:

6. L.H.S. = cos (\(\frac{π}{4}\) – x)cos(\(\frac{π}{4}\) – y) – sin (\(\frac{π}{4}\) – x)sin(\(\frac{π}{4}\) – y)
Put \(\frac{π}{4}\) – x = A and \(\frac{π}{4}\) – y = B, we get:
L.H.S. = cos A cos B – sin A sin B
= cos (A + B)
= cos[(\(\frac{π}{4}\) – x) + (\(\frac{π}{4}\) – y)] = cos [\(\frac{π}{2}\) – (x + y)] [∵cos (\(\frac{π}{2}\) – θ) = sin θ]
= sin (x + y) = R.H.S.

7.
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 14

8. L.H.S. = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)
Now, cos (π + x) = – cos x, cos (- x) = cos x,
sin (π – x) = sin x, cos (\(\frac{π}{2}\) + x) = – sin x
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 4
= cot2x = R.H.S.

9. L.H.S. = cos (\(\frac{3π}{2}\) + x) cos (2π + x) [cot (\(\frac{3π}{2}\) – x) + cot (2π + x)]
Now, cos (\(\frac{3π}{2}\) + x) = sin x, cos (2π + x) = cos x,
cot (\(\frac{3π}{2}\) – x) = tan x, cot (2π + x) = cot x.
∴ L.H.S. = sin x cos x [(tan x) + cot x]
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 5
= 1 = R.H.S.

GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Prove the following:
10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
11. cos (\(\frac{3π}{4}\) + x) – cos (\(\frac{3π}{4}\) – x) = – \(\sqrt{2}\) sin x
12. sin26x – sin24x = sin 2x sin 8x
13. cos22x – cos26x = sin 4x sin 8x
14. sin 2x + 2 sin 4x + sin 6x = 4 cos2x sin 4x
15. cot 4x(sin 5x + sin 3x) = cot x(sin 5x – sin 3x)
Solutions to questions 10 – 15:
10. L.H.S. = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x
= cos (n + 2)x cos (n + 1)x + sin (n + 2)x sin (n + 1)x
Put (n + 2)x = A, (n + 1)x = B.
L.H.S. = cos A cos B + sin A sin B
= cos (A – B)
= cos [(n + 2)x – (n + 1)x]
= cos x = R.H.S.

11. L.H.S. = cos(\(\frac{3π}{4}\) + x) – cos(\(\frac{3π}{4}\) – x)
Put \(\frac{3π}{4}\) + x = θ, \(\frac{3π}{4}\) – x = ϕ.
∴ L.H.S. = cos θ – cos ϕ
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 6

12. L.H.S. = sin2 6x – sin2 4x
We apply the formula sin2A – sin2B = sin (A + B) sin (A – B).
= (sin A cos B + cos A sin B)(sin A cos B – cos A sin B)
= sin2A cos2B – cos2A sin2B
= sin2A (1 – sin2B) – (1 – sin2 A) sin2B
= sin2A – sin2A sin2B – sin2 B + sin2A sin2B
= sin2A – sin2B
∴ L.H.S. = sin26x – sin2 4x = sin (6x + 4x) sin (6x – 4x)
= sin 10x sin 2x = R.H.S.

13. L.H.S. = cos22x – cos2 6x = 1 – sin2 2x – (1 – sin2 6x)
= sin2 6x – sin2 2x = sin (6x + 2x)sin (6x – 2x)
= sin 8x sin 4x = R.H.S.

14. L.H.S. = sin 2x + 2sin 4x + sin 6x
= (sin 6x + sin 2x) + 2 sin 4x
= 2 sin \(\frac{6x+2x}{2}\) cos \(\frac{6x-2x}{2}\) + 2sin 4x
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 2sin 4x cos 2x + 2sin 4x
= 2sin 4x [cos 2x + 1]
= 2 sin 4x 2cos2x
[∵ cos 2x + 1 = 2cos2x – 1 + 1 = 2cos2 x]
= 2sin 4x 2cos2x
= 4sin 4x cos2x = R.H.S.

15. L.H.S. = cot 4x [sin 5x + sin 3x]
= cot 4x [2sin 4x cosx]
= \(\frac{cos x}{sin x}\) × 2sin 4x cos x
= 2cosx cos 4x
= \(\frac{cos x}{sin x}\) × (2cos 4x sin x)
= \(\frac{cos x}{sin x}\) × (sin 5x – sin 3x) = cot x (sin 5x – sin 3x)

GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Prove the following:
16. \(\frac{cos 9x – cos 5x}{sin 17x – sin 3x}\) = – \(\frac{sin x}{cos 10x}\)
17. \(\frac{sin 5x + sin 3x}{cos 5x + cos 3x}\) = tan 4x
18. \(\frac{sin x – sin y}{cos x + cos y}\) = tan \(\frac{x – y}{2}\)
19. \(\frac{sin x + sin 3x}{cos x + cos 3x}\) = tan 2x
20. \(\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}\) = 2 sin x
Solutions to questions 16 – 20:
16. L.H.S. =
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 7

17. L.H.S. =
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 8
= \(\frac{sin 4x}{cos 4x}\) = tan 4x = R.H.S.

18. L.H.S. =
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 9

19. L.H.S. =
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 10

20. L.H.S.=
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 11

GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Prove the following:
21. \(\frac{cos 4x + cos 3x + cos 2x}{sin 4x + sin 3x + sin 2x}\) = cot 3x
22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
23. tan 4x = \(\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}\)
24. cos 4x = 1 – 8 sin2x cos2 x
25. cos 6x = 32 cos6x = 48cos4x + 18cos2x – 1
Solutions to questions 21 – 25:
21. L.H.S. =
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 12

22. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
We have: 3x = x + 2x
cot 3x = cot (x + 2x)
= \(\frac{cot xcot2x – 1}{cot x + cot 2x}\)
By cross multiplication,
cot 3x (cot x + cot 2x) = cot x cot 2x – 1
cot x cot 3x + cot 2x cot 3x = cos x cot 2x – 1
∴ cos x cot 2x – cot 2x cot 3x – cot 3x cot x = 1 = R.H.S.

23. L.H.S. = tan 4x = tan 2(2x)
GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3 img 13

24. L.H.S. = cos 4x = cos 2(2x)
= 2cos2 2x – 1 [∵ cos 2A = 2cos2A – 1]
= 2[2 cos2 x – 1]2 – 1
= 2[4 cos4x – 4 cos2x + 1] – 1
= 8cos4x – 8cos2x + 2 – 1
= 1 – 8 cos2x (1 – cos2x) = 1 – 8cos2xsin2x = R.H.S.

25. cos 6x = cos 3(2x) = 4 cos32x – 3cos 2x
Putting cos 2x = 2cos2x – 1, we get [∵ cos 3A = cos (2A + A) = cos 2AcosA – sin 2Asin A]
= (2cos2A – 1)cos A – 2cos Asin2 A
= 2cos3A – cos A – 2cos A(1 – cos2 A)
= 4 cos3A – 3 cos A
∴ cos 6x = 4(2cos2x – 1)3 – 3(2cos2 x – 1)
= 4(8 cos6x – 12cos4x + 6cos2 x – 1) – 6cos2x + 3
= 32cos6x – 48cos4x + 18cos2x – 1 = R.H.S.

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