#### Answer

$2\tan x\csc 2x-\tan^2 x=1$

#### Work Step by Step

Start with the left side:
$2\tan x\csc 2x-\tan^2 x$
Simplify:
$=2*\frac{\sin x}{\cos x}*\frac{1}{\sin 2x}-\tan^2 x$
$=2*\frac{\sin x}{\cos x}*\frac{1}{2\sin x\cos x}-\tan^2 x$
$=\frac{2\sin x}{2\sin x\cos^2 x}-\tan^2 x$
$=\frac{1}{\cos^2 x}-\tan^2 x$
$=\sec^2 x-\tan^2 x$
$=1$
Since this equals the right side, the identity has been proven.