Answer
$\frac{2~tan~\theta~cos^2~\theta-tan~\theta}{1-tan^2~\theta} = tan~\theta~cos^2~\theta$
Work Step by Step
$\frac{2~tan~\theta~cos^2~\theta-tan~\theta}{1-tan^2~\theta}$
$= (tan~\theta)~(\frac{2~cos^2~\theta-1}{1-tan^2~\theta})$
$= (tan~\theta)~\frac{2~cos^2~\theta-(sin^2~\theta+cos^2~\theta)}{\frac{cos^2~\theta-sin^2~\theta}{cos^2~\theta}}$
$= (tan~\theta)~\frac{(cos^2~\theta-sin^2~\theta)(cos^2~\theta)}{cos^2~\theta-sin^2~\theta}$
$= tan~\theta~cos^2~\theta$
