Answer
$\frac{2\tan x}{1+\tan^2 x}=\sin 2x$
Work Step by Step
Start from the left side:
$\frac{2\tan x}{1+\tan^2 x}$
Rewrite everything in terms of sine and cosine:
$=\frac{2*\frac{\sin x}{\cos x}}{1+\frac{\sin^2 x}{\cos ^2 x}}$
Multiply top and bottom by \cos ^2 x:
$=\frac{(2*\frac{\sin x}{\cos x})\cos ^2 x}{(1+\frac{\sin^2 x}{\cos ^2 x})\cos ^2 x}$
Simplify:
$=\frac{2\sin x\cos x}{\cos^2x+\sin^2x}$
Use the identities $2\sin x\cos x=\sin 2x$ and $\cos^2x+\sin^2x=1$:
$=\frac{\sin 2x}{1}$
$=\sin 2x$
Since this equals the right side, the identity is proved.
