#### Answer

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

#### Work Step by Step

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