NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3

Chapter 3 Trigonometric Functions Exercise 3.3 NCERT Solutions for Class 11 is helpful in clearing your concepts about the chapter and solving your doubts easily. Class 11 Maths NCERT Solutions will be useful in building your basic concepts and preparing for higher classes. The NCERT solutions provided here are prepared by Studyrankers subject matter experts that will guide whenever you got problem while solving a question.

1. Prove that:

Sin² π/6 + cos² π/3 –tan² π/4 = -1/2h

Answer

L.HS. = sin² π/6 + cos² π/3 – tan² π/4

= (sin π/6)² + (cos π/3)² – (tan π/4)²

=(1/2)² + (1/2)² – 1²

[sin π/6 = ½, cos π/3 = 1/2, tan π/4 = 1]

= 1/4 + 3/4 - 1 = 1/2 - 1 = -1/2 = R.H.S

2. 2sin2 π/6 + cosec2 7 π/6 cos2 π/3 = 3/2

Answer

L.H.S = 2sin² π/6 + cosec² π/3 cos² 7π/6

= 2(sin π/6)² + (cosec 7π/6)² (cos π/3)²

= 2.(1/2)² + [cosec(π + π/6)]² (1/2)²

= 2 × 1/4+ (-cosec π/6)² (1/4)

[cosec (π +θ ) = - cosec]

= 1/2 + (2)² 1/4= 1/2 + 1 = 3/2 = R.H.S

3. Cot² π/6 + cosec 5π/6 + 3tan² π/6 = 6

Answer

L.H.S = cot² π/6 + cosec 5π/6 + 3tan² π/6
= (cot π/6)² + cosec(π – π/6) + 3(tan π/6)²
=(√3)² + cosec π/6 + 3 (1/√3)²
= 3 + 2 + 3 × 1/3 = 3 + 2 + 1 = 6 = R.H.S.

4. 2sin² 3π/4 + 2cos² π/4 + 2sec² π/3 = 10

Answer

L.H.S = 2sin² 3π/4 + 2cos² π/4 + 2sec² π/3
= 2(sin 3π/4)² + 2(cos π/4)² + 2(sec π/3)²
= 2(sin [π – π/4])² + 2 × (1/2)² + 2(2²)
= 2(sin π/4)² + 2 × 1/2 + 8
= 2 × (1/2)² + 1 + 8 = 2 × 1/2 + 1 +8
= 1 + 1 + 8 = 10 = R.H.S

5. Find the value of :
(i) sin 75°
(ii) tan 15°

Answer

(i) We have

sin 75° = sin(45° + 30°)

=sin45° cos 30° + cos 45° sin30°

= (1/√2)(√3/2) + (1/√2)(1/2)

= √3/2√2 + 1/2√2 = (√3 + 1)/2√2

= {(√3 + 1)√2}/4 = 1/4 (√2 + √6)

(ii) similar tan 15° = tan(45° - 30°)

= (tan 45° - tan 30°)/(1 + tan 45° tan 30°)

= (1 - 1/√3)/{1 + (1)(1/√3) = {(√3 -1)/√3}/{(√3 + 1)/√3}

= (√3 - 1)/(√3 + 1) = (√3 - 1)/√3 + 1) × (√3 - 1)/(√3 - 1)

= {(√3)² - 2√3 + (1)²}/(3 - 1)

= (3 - 2√3 + 1)/2 = (4 -2√3)/2

={2(2 - √3)}/2 = 2 -√3

6. Prove that:
Cos(π/4 – x)cos(π/4 – y) – sin(π/4 – x) sin(π/4 – y) = sin( x + y)

Answer

L.H.S = cos (π/4 – x) cos(π/4 – y) – sin(π/4 – x) sin(π/4 – y)
Let π/4 – x = A, π/4 – y = B
L.H.S = cos A cosB – sin A sin B
= cos (A + B)
= cos(π/4 – x + π/4 – y)
= cos(π/2 – (x + y))
= sin( x + y) = R.H.S

7. Prove that {tan(π/4 + x)}/{tan(π/4 – x)} = {(1 + tan x)/(1 – tan x)}2

Answer

L.H.S = {tan(π/4 + x)}/{tan(π/4 – x)} = {(1 + tan x)/(1 – tan x)}/{(1 – tan x)/(1 + tan x)}
= {(1 + tan x)/( 1 – tan x)}2 = R.H.S

8. Prove that {cos(π + x)cos(-x)}/{sin( π -x)cos(π/2 + x)} = cot²x

Answer

L.H.S = {cos( π + x)cos(-x)}/{sin(π -x)cos(π/2 + x)}
= {(-cosx)cosx/{sinx(-sinx)} = -cos²x/-sin²x
= cot²x = R.H.S

9. Prove that,
Cos(3π/2 + x)cos(2π + x)[cot(3π/2 – x) + cos(2π + x)] = 1

Answer

L.H.S = cos(3π/2 + x) cos(2π + x)
[cot(3π/2 – x) + cot(2π + x)
Now,
Cos(3π/2 + x) = sin x, cos(2π + x) = cos x and
Cot(3π/2 - x) = tanx, cot(2π + x) = cot x
L.H.S = sin x. cos x[tan x + cot x]
= sin x.cos x[sinx/cos x + cosx/sinx]
= sin x. cos x[(sin² x + cos²x)/cosxsinx]
= (sin x. cosx)1/cosxsinx = 1
[∵ sin²x + cos²x = 1]

10. Prove that sin( n + 1)x sin( n + 2)x + cos(n + 1)x cos(n + 1)x cos(n + 2)x = cosx.

Answer

L.H.S = sin( n+ 1)x sin(n + 2)x + cos( n+1)x cos( n + 2)x
Write A = (n + 1)x
B = ( n + 2)x
L.H.S = sin A sin B + cos A cos B
= cos (A – B)
=cos ((n + 1)x – (n + 2)x)
= cos( x – 2x)
= cos(-x) [ cos(-θ) = cosθ ∀ θ ∈ R]
= cos x = R.H.S

11. Prove that,
Cos(3π/4 + x) – cos(3π/4 – x) = - √2 sin x

Answer

L.H.S = cos(3π/4 + x) – cos(3π/4 – x),
= -2sin ((3π/4 + x + 3π/4 - x)/2)sin((3π/4 + x – 3π/4 + x)/2)
= - 2sin ((3π/2))/2 sin(2x/2) = -2sin 3π/4 sin x
= -2sin(π - π/4)sin x = -2 sin π/4 sin x
= -2 × 1/√2 sin x = - √2sin x = R.H.S

12. Prove that
Sin²6x –sin²4x = sin2x sin10x

Answer

L.H.S = sin² 6x – sin²4x
= (1 – cos 12x)/2 – (1 – cos 8x)/2
= (cos 8x – cos12x)/2
= (2 sin 10x. sin 2x)/2
=sin 10x . sin 2x = R.H.S Proved

13. Prove that
Cos²2x – cos²6x = sin 4x sin 8x

Answer

L.H.S = cos²2x – cos²6x
= (cos 2x + cos 6x)(cos 2x – cos 6x)

= [(2cos 4x cos(-2x))(-2 sin 4x sin (-2x))]
=(2 cos 4x xos 2x)(2 sin 4x sin 2x)
[ cos(-θ) = θ , sin(-θ) = -sinθ]
(2 sin 4x cos 4x)(2 sin 2x cos 2x)
= sin 8x sin 4x = R.H.S
[ sin 2A = 2 sin A cos A]

14. Prove that
Sin26x –sin24x = sin2x sin10x

Answer

L.H.S = sin2x + 2 sin 4x + sin 6x
= (sin 6x + sin 2x) + 2sin4x
(sin C + sin D = 2 sin (C + D)/2 cos (C – D)/2)
= 2sin (6x + 2x)/2 cos (6x – 2x)/2 + 2 sin 4x
= 2 sin 4x [cos 2x + 1]
[ cos 2x + 1 = 2 cos²x – 1 + 1 = 2 cos²x]
= 2 sin 4x cos²x
= 4 sin 4x cos²x = R.H.S.

15. Prove that,
Cot 4x[sin 5x + sin 3x] = cotx(sin 5x – sin 3x)

Answer

L.H.S = cot4x[sin 5x + sin 3x]
= cot 4x[2sin 4x cos x]
=cos4x/sin4x × 2sin 4x cosx = 2cos x cos 4x
= cosx/sinx × (sin5x – sin 3x)
[ ∵ 2cos A sin B = sin(A + B) – sin (A – B)]
= cotx (sin 5x – sin 3x) = R.H.S

16. Prove that (cos 9x – cox 5x)/(sin 17x – sin 3x) = - sin 2x/cos 10x

Answer

L.H.S = (cos 9x – cox 5x)/(sin 17x – sin 3x)
= {-2sin (9x + 5x/2) sin(9x – 5x/2)}/{2cos(17x + 3x/2)sin(17x – 3x)/2}

= (-sin 7x .sin 2x)/(cos 10x. sin 7x) = - sin 2x/cos 10x = R.H.S

17. Prove that (sin5x + sin 3x)/(cos5x + cos 3x) = tan 4x

Answer

L.H.S
= (sin5x + sin 3x)/(cos5x + cos 3x) = (2sin(5x + 3x/2)cos(5x – 3x/2))/(2cos(5x +3x/2)cos(5x -3x/2))
= sin4x/cos4x = tan 4x = R.H.S

18. Prove that (sin x – sin y)/(cos x + cos y) = tan (x – y)/2

Answer

L.H.S = (sin x – sin y)/(cos x + cos y) = (2cos (x + y/2)sin(x – y/2))/(2cos(x + y/2)cos(x – y/2))
= sin(x – y/2)/cos(x – y/2) = tan (x – y)/2 = R.H.S

19. Prove that (sin x + sin 3x)/(cos x + cos 3x) = tan2x

Answer

L.H.S = (sin x + sin 3x)/(cos x + cos 3x) = (sin 3x + sin x)/(cos3x + cosx)
= {2sin(3x + x/2) cos(3x – x/2)}/{2cos(3x + x/2)cos(3x – x/2)} = (sin2xcosx)/(cos2xcosx)
=sin2x/cos2x = tan 2x

20. Prove that (sinx – sin3x)/sin²x – cos²x = 2sin x

Answer

L.H.S = sinx – sin 3x)/sin²x – cos²x = -(sin 3x – sin x)/-(cos²x – sin²x)
= (sin 3x – sin x)/cos²x – sin²x = {2cos(3x + x/2)sin(3x –x/2)}/cos2x
= (2cos2x × sinx)/cos2x [ ∵ cos 2x = cos²x – sin²x]
= 2sin x

21. Prove that (cos 4x + cos 3x + cos 2x)/(sin4x + sin 3x + sin 2x) = cot 3x

Answer

L.H.S = (cos4x + cos3x + cos2x)/(sin4x + sin 3x + sin 2x)
= (cos 4x + cos 2x + cos 3x)/(sin 4x + sin2x + sin 3x)
= {2cos (4x + 2x/2)cos(4x – 2x/2) + cos3x}/{2 sin(4x +2x/2)cos(4x – 2x/2) + sin 3x}
= (2cos3xcosx + cos 3x)/(2sin 3x cos x + sin 3x)
= cos3x(2cosx + 1)/sin3x(2cosx + 1)
= cos3x/sin 3x = cot 3x = R.H.S

22. cot x cot 2x – cot2x cot 3x – cot 3x cot x = 1

Answer

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)
= (cot x cot 2x – 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 cot3x – cot 3x cot x = 1

23. Prove that tan 4x = (4tan x( 1 – tan²x)/(1- 6tan² x + tan⁴x)

Answer

L.H.S = tan 4x = tan2(2x)
= 2tan2x/1 - tan²2x = 2.(2tan x/1 – tan²x)/1 – (2tan x/1 – tan²x)²


4tan x/(1 – tan²x) ( 1 – 6tan²x + tan⁴x) = RHS

24. Prove that cos 4x = 1 - 8 sin²x cos²x

Answer

L.H.S = cos4x = cos2(2x)
= 2cos²2x – 1 [cos 2A = 2cos² A -1]
= 2[2cos² x – 1]² – 1
= 2[4cos⁴x – 4cos² x + 1] – 1
= 8cos⁴x – 8cos²x + 2 – 1
= 1 – 8cos²x(1 – cos²x)
= 1 – 8 cos²x sin²x = R.H.S

25. Prove that
Cos6x = 32cos⁶x – 48cosx⁴ + 18cos²x – 1

Answer

Cos6x = cos3(2x) = 4cos³2x – 3cos2x
Put cos2x = 2cos²x -1
Cos6x = 4(2cos²x – 1)³ – 3(2cos²x – 1)
= 4(8cos⁶x – 12cosx⁴x + 6cos²x – 1) – 6cos²x + 3
= 32cos⁶x – 48cos⁴x + 18cos²x – 1 = RHS