Answer
$\cot 2x=\frac{1-\tan^2 x}{2 \tan x}$
Work Step by Step
Start with the left side:
$\cot 2x$
Rewrite cotangent as the reciprocal of tangent:
$=\frac{1}{\tan 2x}$
Use the identity $\tan 2x=\frac{2 \tan x}{1-\tan^2 x}$:
$=\frac{1}{\frac{2 \tan x}{1-\tan^2 x}}$
Multiply top and bottom by $1-\tan^2 x$:
$=\frac{1*(1-\tan^2 x)}{\frac{2 \tan x}{1-\tan^2 x}*(1-\tan^2 x)}$
Simplify:
$=\frac{1-\tan^2 x}{2 \tan x}$
Since this equals the right side, the identity has been proven.
