#### Answer

$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}=\cot x$

#### Work Step by Step

Start with the left side:
$\frac{1}{2}\cot\frac{x}{2}-\frac{1}{2}\tan\frac{x}{2}$
Factor out $\frac{1}{2}$:
$=\frac{1}{2}(\cot\frac{x}{2}-\tan\frac{x}{2})$
Write cotangent as 1 divided by tangent:
$=\frac{1}{2}(\frac{1}{\tan\frac{x}{2}}-\tan\frac{x}{2})$
Use half-angle identities for tangent:
$=\frac{1}{2}\left(\frac{1}{\frac{\sin x}{1+\cos x}}-\frac{1-\cos x}{\sin x}\right)$
Simplify:
$=\frac{1}{2}(\frac{1+\cos x}{\sin x}-\frac{1-\cos x}{\sin x})$
$=\frac{1}{2}*\frac{(1+\cos x)-(1-\cos x)}{\sin x}$
$=\frac{1}{2}*\frac{1+\cos x-1+\cos x}{\sin x}$
$=\frac{1}{2}*\frac{2\cos x}{\sin x}$
$=\frac{\cos x}{\sin x}$
$=\cot x$
Since this equals the right side, the identity has been proven.