#### Answer

$2\cos^2 \theta-1=\frac{1-\tan^2\theta}{1+\tan^2\theta}$

#### Work Step by Step

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.