## Precalculus (6th Edition)

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