## Precalculus (6th Edition)

$\frac{1-\cos x}{1+\cos x}=(\cot x-\csc x)^2$
Start with the left side: $\frac{1-\cos x}{1+\cos x}$ Multiply top and bottom by $1-\cos x$: $=\frac{1-\cos x}{1+\cos x}*\frac{1-\cos x}{1-\cos x}$ $=\frac{1-2\cos x+\cos^2 x}{1-\cos^2x}$ $=\frac{1-2\cos x+\cos^2 x}{\sin^2x}$ $=\frac{1}{\sin^2x}-\frac{2\cos x}{\sin^2x}+\frac{\cos^2 x}{\sin^2x}$ $=\csc^2 x-2*\frac{\cos x}{\sin x}*\frac{1}{\sin x}+\cot^2 x$ $=\csc^2 x-2\cot x\csc x+\cot^2 x$ $=\cot^2 x-2\cot x\csc x+\csc^2 x$ $=(\cot x-\csc x)^2$ Since this equals the right side, the equation has been proven.