Answer
$\frac{\tan x}{1-\cos x}=\csc x(1+\sec x)$
Work Step by Step
Start with the right side:
$\csc x(1+\sec x)$
Multiply top and bottom by $1-\cos x$:
$=\frac{\csc x(1+\sec x)(1-\cos x)}{1-\cos x}$
Multiply $1+\sec x$ by $1-\cos x$ and simplify:
$=\frac{\csc x(1-\cos x+\sec x-\sec x\cos x)}{1-\cos x}$
$=\frac{\csc x(1-\cos x+\sec x-1)}{1-\cos x}$
$=\frac{\csc x(\sec x-\cos x)}{1-\cos x}$
$=\frac{\csc x\sec x-\csc x\cos x}{1-\cos x}$
Write everything in terms of sine and cosine and simplify:
$=\frac{\frac{1}{\sin x}*\frac{1}{\cos x}-\frac{1}{\sin x}*\cos x}{1-\cos x}$
$=\frac{\frac{1}{\sin x\cos x}-\frac{\cos x}{\sin x}}{1-\cos x}$
$=\frac{\frac{1}{\sin x\cos x}-\frac{\cos x}{\sin x}*\frac{\cos x}{\cos x}}{1-\cos x}$
$=\frac{\frac{1}{\sin x\cos x}-\frac{\cos^2 x}{\sin x\cos x}}{1-\cos x}$
$=\frac{\frac{1-\cos^2 x}{\sin x\cos x}}{1-\cos x}$
Simplify using the identity $1-\cos^2 x=\sin^2 x$:
$=\frac{\frac{\sin^2 x}{\sin x\cos x}}{1-\cos x}$
$=\frac{\frac{\sin x}{\cos x}}{1-\cos x}$
$=\frac{\tan x}{1-\cos x}$
Since this equals the left side, the identity has been proven.
