#### Answer

$\sin^2 \frac{x}{2}=\frac{\tan x-\sin x}{2\tan x}$

#### Work Step by Step

Start with the left side:
$\sin^2 \frac{x}{2}$
Use the identity $\sin \frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}$:
$=\left(\pm\sqrt{\frac{1-\cos x}{2}}\right)^2$
$=\frac{1-\cos x}{2}$
Multiply top and bottom by $\tan x$:
$=\frac{1-\cos x}{2}*\frac{\tan x}{\tan x}$
$=\frac{\tan x-\cos x\tan x}{2\tan x}$
$=\frac{\tan x-\cos x*\frac{\sin x}{\cos x}}{2\tan x}$
$=\frac{\tan x-\sin x}{2\tan x}$
Since this equals the right side, the identity has been proven.