#### Answer

$\sec^2\alpha-1=\frac{\sec 2\alpha-1}{\sec 2\alpha+1}$

#### Work Step by Step

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.