#### Answer

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

#### Work Step by Step

Simplify the left side:
$-\cot \frac{x}{2}$
Write cotangent as 1 divided by tangent:
$=-\frac{1}{\tan \frac{x}{2}}$
Simplify:
$=-\frac{1}{\frac{1-\cos x}{\sin x}}$
$=-\frac{\sin x}{1-\cos x}$
$=\frac{\sin x}{\cos x-1}$
Simplify the right side:
$\frac{\sin 2x+\sin x}{\cos 2x-\cos x}$
Expand using the double-angle identities:
$=\frac{2\sin x\cos x+\sin x}{2\cos^2-1-\cos x}$
$=\frac{2\sin x\cos x+\sin x}{2\cos^2-\cos x-1}$
Factor:
$=\frac{\sin x(2\cos x+1)}{(2\cos x+1)(\cos x-1)}$
$=\frac{\sin x}{\cos x-1}$
Since the left side and the right side both simplify to $\frac{\sin x}{\cos x-1}$, they are equal to each other, and the identity has been proven.