#### Answer

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

#### Work Step by Step

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