## Precalculus (6th Edition)

$\cot 4\theta=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$
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 $1-\tan^2 2\theta$: $=\frac{1}{\frac{2\tan 2\theta}{1-\tan^2 2\theta}}*\frac{1-\tan^2 2\theta}{1-\tan^2 2\theta}$ $=\frac{1-\tan^2 2\theta}{2\tan 2\theta}$