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}+xright)}{tan left(frac{pi}{4}-xright)}=frac{(1+tan x)^{2}}{(1-tan x)^{2}})
8. (frac{cos (pi+x) cos (-x)}{sin (pi-x) cos left(frac{pi}{2}+xright)}) = 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}+xright)})
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 xleft(1-tan ^{2} xright)}{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.