## Precalculus (6th Edition)

$\frac{2\cot x}{\tan 2x}=\csc^2 x-2$
Start with the left side: $\frac{2\cot x}{\tan 2x}$ Simplify using the Double-Angle Formula for tangent: $=\frac{2\cot x}{\frac{2\tan x}{1-\tan^2 x}}$ Multiply the top and bottom by $1-\tan^2 x$: $=\frac{2\cot x}{\frac{2\tan x}{1-\tan^2 x}}*\frac{1-\tan^2 x}{1-\tan^2 x}$ Simplify: $=\frac{2\cot x(1-\tan^2 x)}{2\tan x}$ $=\frac{\cot x(1-\tan^2 x)}{\frac{1}{\cot x}}$ Multiply the top and bottom by $\cot x$: $=\frac{\cot x(1-\tan^2 x)}{\frac{1}{\cot x}}*\frac{\cot x}{\cot x}$ Simplify: $=\cot^2 x(1-\tan^2 x)$ $=\cot^2 x-1$ $=(\csc^2 x-1)-1$ $=\csc^2 x-2$ Since this equals the right side, the identity has been proven.