## Precalculus (6th Edition)

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