Answer
$1+tan^2~\alpha = 2~tan~\alpha ~csc~2\alpha$
Work Step by Step
$1+tan^2~\alpha = 1+\frac{sin^2~\alpha}{cos^2~\alpha}$
$1+tan^2~\alpha = \frac{cos^2~\alpha}{cos^2~\alpha}+\frac{sin^2~\alpha}{cos^2~\alpha}$
$1+tan^2~\alpha = \frac{cos^2~\alpha + sin^2~\alpha}{cos^2~\alpha}$
$1+tan^2~\alpha = \frac{1}{cos^2~\alpha}$
$1+tan^2~\alpha = \frac{2~sin~\alpha}{2~sin~\alpha~cos^2~\alpha}$
$1+tan^2~\alpha = \frac{2~sin~\alpha}{2~sin~\alpha~cos~\alpha~cos~\alpha}$
$1+tan^2~\alpha = \frac{2~sin~\alpha}{sin~2\alpha~cos~\alpha}$
$1+tan^2~\alpha = 2~tan~\alpha ~csc~2\alpha$
