## Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3

Question 1.

Prove that

.

Solution:

Question 2.

.

Solution:

Question 3.

.

Solution:

Question 4.

.

Solution:

Question 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)

Question 6.

Prove that:

tan 225° cot 405° + tan 765° cot 675° = 0.

Solution :

L.H.S.

= tan 225° cot 405° + tan 765° cot 675°

= tan (360° – 135°) cot (360° + 45°) + tan (2 × 360° + 45°) cot (2 × 360° – 45°)

= (- tan 135°) (cot 45°) + (tan 45°) (- cot 45°)

= – tan 135° cot 45° – tan 45° cot 45°

= – tan (180° – 45°) cot 45° – tan 45° cot 45°

= – (- tan 45°) cot 45° – tan 45° cot 45°

= tan 45° cot 45° – tan 45° cot 45° = 0

= R.H.S.

Hence Proved.

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

sin (n + 1) x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution :

L.H.S.

= sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

= cos (n + 1)x cos (n + 2)x + sin (n + 1)x × sin (n + 2)x

[Formula cos (x – y) – cos x cos y + sin x sin y]

= cos [(n + 1)x – (n + 2)x]

= cos [nx + x – nx – 2x]

= cos (- x) cos x = R.H.S.

Hence Proved.

Question 12.

sin^{2} 6x – sin^{2} 4x = sin 2x sin 10x.

Solution :

L.H.S. = sin^{2} 6x – sin^{2} 4x

= [2 sin^{2} 6x – 2sin^{2} 4x] [Multiply and divide by 2]

= [(1 – cos 12x) – (1 – cos 8x)]

= sin 10x sin 2x =sin 2x sin 10x = R.H.S.

**Second Method: R.H.S.**

= sin^{2} 6x – sin^{2} 4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (2 sin 5x cos 5x).(2 sin x cos x)

= sin 10x.sin 2x [∵ sin 2θ = 2 sin θ cos θ]

= sin 2x sin 10x = R.H.S. Hence Proved.

Question 13.

sin 2x + 2 sin 4x + sin 6x = 4 cos^{2} x sin 4x.

Solution :

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= (sin 2x + sin 6x) + 2 sin 4x

Question 14.

cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Solution :

L.H.S. = cot 4x (sin 5x + sin 3x)

From (i) and (ii), we get

cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).

L.H.S. = R.H.S.

Hence Proved.

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

cos 4x = 1 – 8 sin^{2} x cos^{2 }x.

Solution :

L.H.S. = cos 4x = cos 2(2x)

(cos θ = 1 – 2 sin^{2} θ)

= 1 – 2 sin^{2} 2x

= 1 – 2 (2 sin x cos x)^{2 }[∵ sin 2x = 2 sin x cos x]

= 1 – 2 x 4 sin^{2} x cos^{2} x

= 1 – 8 sin^{2} x cos^{2} x = R.H.S.

Hence Proved.

Question 21.

cos 6x = 32 cos^{6} x- 48 cos^{4} x + 18 cos^{2} x- 1.

Solution :

L.H.S. = cos 6x = cos 3(2x) [∵ cos 3x = 4 cos^{3} x – 3 cos x]

= 4 cos^{3} 2x – 3 cos 2x

= cos 2x (4 cos^{2} 2x – 3)

= cos 2x [4(cos 2x)^{2} – 3] [∵ cos 2x = 2 cos^{2} x – 1]

= cos 2x [4(2 cos^{2} x – l)^{2} – 3]

= cos 2x [4(4 cos^{4} x + 1 – 4 cos^{2} x) – 3]

= cos 2x (16 cos^{4} x + 4 – 16 cos^{2} x – 3)

= (2 cos^{2} x – 1) (16 cos^{4} x + 4 – 16 cos^{2} x – 3)

= 32 cos^{6} x + 8 cos^{2} x – 32 cos^{4} x – 6 cos^{2} x – 16 cos^{4} x – 4 + 16 cos^{2} x + 3

= 32 cos^{6} x – 48 cos^{4} x + 18 cos^{2} x – 1

= R.H.S. Hence Proved.

Question 22.

[1 + cot θ – sec (θ + π/(2)]

[1 + cot θ + sec (θ + π/(2)] = 2 cot θ

Solution :

L.H.S.

[1 + cot θ – sec (θ + π/(2)]

[1 + cot θ + sec (θ + π/(2)]

= [1 + cot θ + cosec θ] [1+ cot θ – cosec θ]

= (1 + cot)^{2} – cosec^{2}θ

= 1 + cot^{2}θ + 2 cotθ – cosec^{2}θ

= 1 + 2 cot θ – (cosec2θ – cot2θ)

= 1+2 cot θ – 1

= 2 cot θ = R.H.S.

Hence Proved.

##### RBSE Solutions for Class 11 Maths

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