Answer
$\cot 4\theta=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$
Work Step by Step
Start with the left side:
$\cot 4\theta$
Rewrite cotangent as the inverse function of tangent:
$=\frac{1}{\tan 4\theta}$
Rewrite $\tan 4\theta$ as $\tan (2*2\theta)$ and use the double-angle identity:
$=\frac{1}{\tan (2*2\theta)}$
$=\frac{1}{\frac{2\tan 2\theta}{1-\tan^2 2\theta}}$
Multiply top and bottom by $1-\tan^2 2\theta$:
$=\frac{1}{\frac{2\tan 2\theta}{1-\tan^2 2\theta}}*\frac{1-\tan^2 2\theta}{1-\tan^2 2\theta}$
$=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$