Question

sin8x+sin2x/cos2x-cos8x=cot3x

Answer 1

Sin8x + Sin2x/Cos2x - Cos2x
>(2sin(x+y/2)cos(x-y/2))/
(-2sin(x+y/2)sin(x-y/2)

>2sin5xcos3x/-2sin3x(-sin5x)
>2sin5xcos3x/2sin3xsin5x
>Cos2x/sin3x
>Cot3x

Answer 2

Answer and Explanation:

To show :  $\frac{\sin8x+\sin2x}{\cos 2x-\cos 8x} =\cot 3x$

Solution :

Taking LHS,

$\frac{\sin8x+\sin2x}{\cos 2x-\cos 8x}$

Applying formula,

$\sin A+\sin B =2\sin (\frac{A+B}{2})\cos (\frac{A-B}{2})$

$\cos A-\cos B =-2\sin (\frac{A+B}{2})\sin (\frac{A-B}{2})$

Where, A=8x and B = 2x

$=\frac{2\sin (\frac{8x+2x}{2})\cos (\frac{8x-2x}{2})}{-2\sin (\frac{2x+8x}{2})\sin (\frac{2x-8x}{2})}$

$=\frac{2\sin (\frac{10x}{2})\cos (\frac{6x}{2})}{-2\sin (\frac{10x}{2})\sin (\frac{-6x}{2})}$

$=\frac{2\sin (5x)\cos (3x)}{-2\sin (5x)\sin (-3x)}$

We know, $\sin (-\theta)=-\sin \theta$

$=\frac{\cos (3x)}{\sin (3x)}$

$=\cot (3x)$

=RHS

Hence, LHS=RHS