## Precalculus (6th Edition)

$\sec^2\alpha-1=\frac{\sec 2\alpha-1}{\sec 2\alpha+1}$
Simplify the left side: $\sec^2\alpha-1$ $=\tan^2\alpha$ Simplify the right side: $\frac{\sec 2\alpha-1}{\sec 2\alpha+1}$ Rewrite it in terms of cosine: $=\frac{\frac{1}{\cos 2\alpha}-1}{\frac{1}{\cos 2\alpha}+1}$ Multiply top and bottom by $\cos 2\alpha$: $=\frac{\frac{1}{\cos 2\alpha}-1}{\frac{1}{\cos 2\alpha}+1}*\frac{\cos 2\alpha}{\cos 2\alpha}$ $=\frac{1-\cos 2\alpha}{1+\cos 2\alpha}$ Use the identities $\cos 2\alpha=1-2\sin^2\alpha$ and $\cos 2\alpha=2\cos^2\alpha-1$: $=\frac{1-(1-2\sin^2\alpha)}{1+(2\cos^2\alpha-1)}$ Simplify: $=\frac{1-1+2\sin^2\alpha}{1+2\cos^2\alpha-1}$ $=\frac{2\sin^2\alpha}{2\cos^2\alpha}$ $=\tan^2\alpha$ Since the left side and the right side both simplify to $\tan^2\alpha$, they are equal to each other, and the identity has been proven.