Answer
$\frac{2~tan~B}{sin~2B} = sec^2~B$
Work Step by Step
$\frac{2~tan~B}{sin~2B} = \frac{2~\frac{sin~B}{cos~B}}{2~sin~B~cos~B}$
$\frac{2~tan~B}{sin~2B} = \frac{\frac{1}{cos~B}}{cos~B}$
$\frac{2~tan~B}{sin~2B} = \frac{1}{cos^2~B}$
$\frac{2~tan~B}{sin~2B} = sec^2~B$
