#### Answer

$\frac{2}{1+\cos x}-\tan^2 \frac{x}{2}=1$

#### Work Step by Step

Start with the left side:
$\frac{2}{1+\cos x}-\tan^2 \frac{x}{2}$
Use the identity $\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$:
$=\frac{2}{1+\cos x}-\left(\frac{\sin x}{1+\cos x}\right)^2$
$=\frac{2}{1+\cos x}-\frac{\sin^2 x}{(1+\cos x)^2}$
Get a common denominator and subtract:
$=\frac{2}{1+\cos x}*\frac{1+\cos x}{1+\cos x}-\frac{\sin^2 x}{(1+\cos x)^2}$
$=\frac{2(1+\cos x)}{(1+\cos x)^2}-\frac{\sin^2 x}{(1+\cos x)^2}$
$=\frac{2+2\cos x-\sin^2 x}{(1+\cos x)^2}$
Rewrite $\sin^2 x$ as $1-\cos^2 x$:
$=\frac{2+2\cos x-(1-\cos^2 x)}{(1+\cos x)^2}$
Simplify:
$=\frac{2+2\cos x-1+\cos^2 x}{(1+\cos x)^2}$
$=\frac{1+2\cos x+\cos^2 x}{(1+\cos x)^2}$
$=\frac{(1+\cos x)^2}{(1+\cos x)^2}$
$=1$
Since this equals the right side, the identity has been proven.