### Question 15: cot 4x (sin 5x + sin 3x) = cot x (sin 5x – 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 xNow, taking RHS in consideration, we get

= cot x (sin 5x – 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 xHence, 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 2θ = cos^{2}θ – sin^{2}θ=

=

=

=

= 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 – cot 2x cot 3x – cot 3x cot x = 1

**Solution:**

Taking LHS in consideration, we get

= cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x+x) (cot 2x + cot x)

Using the identity,

cot(A+B) == cot x cot 2x –(cot 2x + cot x)

= cot x cot 2x – [cot 2x cot x – 1]

= cot x cot 2x – 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 2θ ==

Again using the same identity, we get

=

=

=

=

=

=

Hence, LHS = RHS

### Question 24: cos 4x = 1 – 8sin^{2}x cos^{2}x

**Solution:**

Taking LHS in consideration, we get

cos 4x = cos 2 (2x)

Using the identity,

cos 2θ = 1 – 2sin^{2}θ= 1 – 2sin

^{2}(2x)= 1 – 2(2sin x cos x)

^{2 }(As,sin 2θ = 2 sin θ cos θ)= 1 – 2(4sin

^{2 }x cos^{2}x)= 1 – 8sin

^{2}x cos^{2}xHence, LHS = RHS

### Question 25: cos 6x = 32 cos^{6}x – 48cos^{4}x + 18 cos^{2}x – 1

**Solution:**

Taking LHS in consideration, we get

cos 6x = cos 3 (2x)

Using the identity,

cos 3θ = 4 cos^{3}θ – 3 cos θ= 4 cos

^{3}(2x) – 3 cos (2x)= 4 cos

^{3}(2x) – 3 cos (2x)Now, Using the identity

cos 2θ = 2cos^{2}θ – 1= 4 (2cos

^{2}x – 1)^{3}– 3 (2cos^{2}x – 1)Using algebraic identity,

(a-b)^{3}= a^{3}+ b^{3}– 3a^{2}b + 3ab^{2}= 4 [(2 cos

^{2}x)^{3}– (1)^{3}– 3 (2 cos^{2}x)^{2}+ 3 (2 cos^{2}x)(1)^{2}] – 6cos^{2}x + 3= 4 [8cos

^{6}x – 1 – 12 cos^{4}x + 6 cos^{2}x] – 6 cos2x + 3= 32 cos

^{6}x – 4 – 48 cos^{4x}+ 24 cos^{2}x – 6 cos^{2}x + 3= 32 cos

^{6}x – 48 cos^{4}x + 18 cos^{2}x – 1Hence, LHS = RHS