Answer
$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$
Work Step by Step
Start with the right side:
$\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$
Use the identity $1+\tan^2 \theta=\sec^2\theta$, where $\theta=\frac{x}{2}$:
$=\frac{1-\tan^2\frac{x}{2}}{\sec^2\frac{x}{2}}$
Rewrite everything in terms of sine and cosine:
$=\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac{1}{\cos^2\frac{x}{2}}}$
Multiply top and bottom by $\cos^2\frac{x}{2}$:
$=\frac{1-\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}}{\frac{1}{\cos^2\frac{x}{2}}}*\frac{\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}}$
$=\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{1}$
$=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$
Use the identity $\cos 2\theta=\cos^2 \theta-\sin^2\theta$, where $\theta=\frac{x}{2}$:
$=\cos (2*\frac{x}{2})$
$=\cos x$
Since this equals the left side, the identity has been proven.