Answer
$\frac{2(\tan x-\cot x)}{\tan^2 x-\cot ^2x}=\sin 2x$
Work Step by Step
Start with the left side:
$\frac{2(\tan x-\cot x)}{\tan^2 x-\cot ^2x}$
Factor the denominator as a difference of perfect squares:
$=\frac{2(\tan x-\cot x)}{(\tan x-\cot x)(\tan x+\cot x)}$
Cancel out $\tan x-\cot x$:
$=\frac{2}{\tan x+\cot x}$
Write everything in terms of sine and cosine:
$=\frac{2}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}$
Multiply top and bottom by $\sin x\cos x$:
$=\frac{2\sin x\cos x}{(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})\sin x\cos x}$
Simplify:
$=\frac{2\sin x\cos x}{\frac{\sin x}{\cos x}*\sin x\cos x+\frac{\cos x}{\sin x}*\sin x\cos x}$
$=\frac{2\sin x\cos x}{\sin^2x+\cos^2x}$
Use the identities $\sin^2x+\cos^2x=1$ and $\sin 2x=2\sin x\cos x$:
$=\frac{\sin 2x}{1}$
$=\sin 2x$
