# Class 11 NCERT Solutions- Chapter 3 Trigonometric Function – Exercise 3.3 | Set 2

Last Updated : 07 Apr, 2021

### Question 15: cot 4x (sin 5x + sin 3x) = cot x (sin 5x â€“ sin 3x)

Solution:

Taking LHS in consideration, we get

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

=(sin 5x + sin 3x)

Using the identity,

sin A + sin B = 2 sincos

=(2 sincos)

=(2 sincos)

=(2 sin 4x cos x)

= 2 cos 4x cos x

Now, taking RHS in consideration, we get

= cot x (sin 5x â€“ sin 3x)

=(sin 5x – sin 3x)

Using the identity,

sin A – sin B = 2 cossin

=(2 cossin)

=(2 cossin)

=(2 cos 4x sin x)

= 2 cos 4x cos x

Hence, LHS = RHS

### Question 16:

Solution:

Taking LHS in consideration, we get

=

Using the identity,

cos A – cos B = 2 sinsin

sin A – sin B = 2 cossin

=

=

=

=

Hence, LHS = RHS

### Question 17:= tan 4x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

=

= tan 4x

Hence, LHS = RHS

### Question 18:

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A – sin B = 2 cossin

cos A + cos B = 2 coscos

=

=

=

Hence, LHS = RHS

### Question 19:= tan 2x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

=

=

= tan 2x

Hence, LHS = RHS

### Question 20:= 2 sin x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A – sin B = 2 cossin

cos 2Î¸ = cos2 Î¸ – sin2 Î¸

=

=

=

=

= 2 sin (x)

Hence, LHS = RHS

### Question 21:= cot 3x

Solution:

Taking LHS in consideration, we get

=

Using the identity,

sin A + sin B = 2 sincos

cos A + cos B = 2 coscos

=

=

=

Taking common, we have

=

=

= cot 3x

Hence, LHS = RHS

### Question 22: cot x cot 2x â€“ cot 2x cot 3x â€“ cot 3x cot x = 1

Solution:

Taking LHS in consideration, we get

= cot x cot 2x â€“ cot 2x cot 3x â€“ cot 3x cot x

= cot x cot 2x â€“ cot 3x (cot 2x + cot x)

= cot x cot 2x â€“ cot (2x+x) (cot 2x + cot x)

Using the identity,

cot(A+B) =

= cot x cot 2x â€“(cot 2x + cot x)

= cot x cot 2x â€“ [cot 2x cot x – 1]

= cot x cot 2x â€“ cot 2x cot x – 1

= 1

Hence, LHS = RHS

### Question 23: tan 4x =

Solution:

Taking LHS in consideration, we get

tan 4x = tan 2(2x)

Using the identity,

tan 2Î¸ =

=

Again using the same identity, we get

=

=

=

=

=

=

Hence, LHS = RHS

### Question 24: cos 4x = 1 â€“ 8sin2x cos2x

Solution:

Taking LHS in consideration, we get

cos 4x = cos 2 (2x)

Using the identity,

cos 2Î¸ = 1 – 2sin2 Î¸

= 1 – 2sin2 (2x)

= 1 – 2(2sin x cos x)2 (As, sin 2Î¸ = 2 sin Î¸ cos Î¸)

= 1 – 2(4sin2 x cos2 x)

= 1 – 8sin2 x cos2 x

Hence, LHS = RHS

### Question 25: cos 6x = 32 cos6x â€“ 48cos4x + 18 cos2x â€“ 1

Solution:

Taking LHS in consideration, we get

cos 6x = cos 3 (2x)

Using the identity,

cos 3Î¸ = 4 cos3 Î¸ â€“ 3 cos Î¸

= 4 cos3 (2x) â€“ 3 cos (2x)

= 4 cos3 (2x) â€“ 3 cos (2x)

Now, Using the identity cos 2Î¸ = 2cos2 Î¸ – 1

= 4 (2cos2 x – 1)3 â€“ 3 (2cos2 x – 1)

Using algebraic identity,

(a-b)3 = a3 + b3 – 3a2b + 3ab2

= 4 [(2 cos2 x) 3 â€“ (1)3 â€“ 3 (2 cos2 x) 2 + 3 (2 cos2 x)(1)2] â€“ 6cos2 x + 3

= 4 [8cos6x â€“ 1 â€“ 12 cos4x + 6 cos2x] â€“ 6 cos2x + 3

= 32 cos6x â€“ 4 â€“ 48 cos4x + 24 cos2 x â€“ 6 cos2x + 3

= 32 cos6x â€“ 48 cos4x + 18 cos2x â€“ 1

Hence, LHS = RHS

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