#### Answer

$\tan^2 x-\sin^2 x=(\tan x\sin x)^2$

#### Work Step by Step

Start with the left side:
$\tan^2 x-\sin^2 x$
Rewrite in terms of sine and cosine:
$=\frac{\sin^2x}{\cos^2 x}-\sin^2 x$
Get a common denominator of $\cos^2 x$:
$=\frac{\sin^2x}{\cos^2 x}-\frac{\sin^2x\cos^2x}{\cos^2 x}$
$=\frac{\sin^2x-\sin^2x\cos^2x}{\cos^2 x}$
Simplify:
$=\frac{\sin^2x(1-\cos^2x)}{\cos^2 x}$
$=\frac{\sin^2x\sin^2x}{\cos^2 x}$
$=\sin^2 x*\frac{\sin^2x}{\cos^2 x}$
$=\sin^2 x\tan^2 x$
$=(\tan x\sin x)^2$
Since this equals the right side, the identity is proven.