#### Answer

$\frac{\sin^2 x-\cos^2 x}{\csc x}=2\sin^3 x-\sin x$

#### Work Step by Step

Simplify the left side:
$\frac{\sin^2 x-\cos^2 x}{\csc x}$
Rewrite it in terms of sine and cosine:
$=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}$
Multiply top and bottom by $\sin x$:
$=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}*\frac{\sin x}{\sin x}$
$=(\sin^2 x-\cos^2x)\sin x$
$=-(\cos^2x-\sin^2 x)\sin x$
Use the identity $\cos^2x-\sin^2 x=\cos 2x$:
$=-\cos 2x\sin x$
Simplify the right side:
$2\sin^3 x-\sin x$
Factor out $\sin x$:
$=\sin x(2\sin^2 x-1)$
$=-\sin x(1-2\sin^2 x)$
Use the identity $1-2\sin^2 x=\cos 2x$:
$=-\sin x\cos 2x$
$=-\cos 2x\sin x$
Since the left side and the right side both simplify to $-\cos 2x\sin x$, they are equal to each other, and the identity has been proven.