Answer
$\frac{2\tan x}{1+\tan^2 x}=\sin 2x$
Work Step by Step
Start with the left 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 everything by $\cos^2 x$:
$=\frac{2*\frac{\sin x}{\cos x}*\cos^2 x}{\frac{1}{\cos^2 x}*\cos^2 x}$
$=\frac{2\sin x\cos x}{1}$
$=2\sin x\cos x$
Use the identity $\sin 2x=2\sin x\cos x$:
$=\sin 2x$
Since this equals the right side, the identity has been proven.
