#### Answer

$\cot^2 \frac{x}{2}=\frac{(1+\cos x)^2}{\sin^2 x}$

#### Work Step by Step

Start with the left side:
$\cot^2 \frac{x}{2}$
Rewrite cotangent as the inverse function of tangent:
$=\frac{1}{\tan^2 \frac{x}{2}}$
Use the half-angle identity $\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}$:
$=\frac{1}{\left(\frac{\sin x}{1+\cos x}\right)^2}$
$=\frac{1}{\frac{\sin^2 x}{(1+\cos x)^2}}$
Multiply top and bottom by $(1+\cos x)^2$:
$=\frac{1}{\frac{\sin^2 x}{(1+\cos x)^2}}*\frac{(1+\cos x)^2}{(1+\cos x)^2}$
$=\frac{(1+\cos x)^2}{\sin^2 x}$
Since this equals the right side, tie identity has been proven.