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 $\frac{π}{3}$

= 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]

(ii) tan 15° = tan (45° – 30°)

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.

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

= 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]

= 1 = R.H.S.

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

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)

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. =

17. L.H.S. =

= $\frac{sin 4x}{cos 4x}$ = tan 4x = R.H.S.

18. L.H.S. =

19. L.H.S. =

20. L.H.S.=

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. =

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)

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.