Answer
$\tan\theta\cos^2\theta=\frac{2\tan\theta\cos^2\theta-\tan \theta}{1-\tan^2\theta}$
Work Step by Step
Start with the right side:
$\frac{2\tan\theta\cos^2\theta-\tan \theta}{1-\tan^2\theta}$
Factor out $\tan \theta$ from the numerator:
$=\frac{\tan\theta(2\cos^2\theta-1)}{1-\tan^2\theta}$
Simplify the denominator:
$=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}$
$=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}$
Multiply top and bottom by $\cos^2 \theta$:
$=\frac{\tan\theta(2\cos^2\theta-1)}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}*\frac{\cos^2\theta}{\cos^2\theta}$
$=\frac{\tan\theta(2\cos^2\theta-1)\cos^2\theta}{\cos^2\theta-\sin^2\theta}$
Use the double-angle identities for cosine:
$=\frac{\tan\theta\cos 2\theta\cos^2\theta}{\cos 2\theta}$
Simplify:
$=\frac{\tan\theta\cos^2\theta}{1}$
$=\tan\theta\cos^2\theta$
Since this equals the left side, the identity has been proven.