Answer
$\frac{1-\cos 2x}{\sin 2x}=\tan x$
Work Step by Step
Start with the left side:
$\frac{1-\cos 2x}{\sin 2x}$
Use the double-angle formulas $\cos 2x=1-2\sin^2 x$ and $\sin 2x=2\sin x\cos x$ to expand the expression:
$=\frac{1-(1-2\sin^2 x)}{2\sin x\cos x}$
Simplify:
$=\frac{1-1+2\sin^2 x}{2\sin x\cos x}$
$=\frac{2\sin^2 x}{2\sin x\cos x}$
$=\frac{\sin x}{\cos x}$
$=\tan x$
Since this equals the right side, the identity is proven.
