Precalculus (6th Edition)

$2\cos^2 \theta-1=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
Start with the right side: $\frac{1-\tan^2\theta}{1+\tan^2\theta}$ Simplify: $=\frac{1-\tan^2\theta}{\sec^2\theta}$ $=\frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}$ $=(1-\frac{\sin^2\theta}{\cos^2\theta})\cos^2\theta$ $=\cos^2\theta-\frac{\sin^2\theta}{\cos^2\theta}\cos^2\theta$ $=\cos^2\theta-\sin^2\theta$ $=\cos 2\theta$ $=2\cos^2\theta-1$ Since this equals the left side, the identity has been proven.