#### Answer

$\frac{1-\cos x}{1+\cos x}=(\cot x-\csc x)^2$

#### Work Step by Step

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.