Question

Cos4x - cos3x + cos2x / sin4x + sin3x + sin 2x = cot 3x

Answer 1

mis-spelt in question..... there should be + sign everywhere

Answer 2

Answer:

L.H.S = R.H.S

Step-by-step explanation:

According to this question

Given that

$\frac{cos 4x + cos 3x + cos 2x}{sin 4x + sin 3x +sin 2x}$ = cot 3x

Let L.H.S

$\frac{cos 4x + cos 3x + cos 2x}{sin 4x + sin 3x +sin 2x}$

Formula,

cos a + cos b = $2 cos\frac{a + b}{2} cos\frac{a - b}{2}$

sin a + sin b = $2 sin\frac{a + b}{2} cos\frac{a - b}{2}$

$\frac{2 cos\frac{4x + 2x}{2} cos\frac{4x - 2x}{2} + cos 3x}{2 sin\frac{4x + 2x}{2} cos\frac{4x - 2x}{2} + sin 3x}$

$\frac{2 cos 3x cos x + cos 3x}{2 sin3x cos x + sin 3x}$

$\frac{cos 3x(1 + 2 cos x)}{sin 3x(1 + 2 cos x)}$

$\frac{cos 3x}{sin 3x}$

cot 3x

R.H.S