#### Answer

$\frac{\tan x-\cot x}{\tan x+\cot x}=2\sin^2 x-1$

#### Work Step by Step

Simplify the left side:
$\frac{\tan x-\cot x}{\tan x+\cot x}$
Rewrite in terms of sine and cosine:
$=\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}$
Multiply top and bottom by $\sin x\cos x$:
$=\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}*\frac{\sin x\cos x}{\sin x\cos x}$
Simplify:
$=\frac{\sin^2x-\cos^2 x}{\sin^2 x+\cos^2 x}$
$=\frac{-(\cos^2 x-\sin^2 x)}{1}$
$=-\cos 2x$
Simplify the right side:
$2\sin^2 x-1$
$=-(1-2\sin^2 x)$
$=-\cos 2x$
Since both the left side and the right side are equal to $=-\cos 2x$, they are equal to each other, and the identity is proven.