Question

Solve the following:

$tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)$

Answer 1

$\begin{array}{c}\frac{d}{dx}\left(\tan \frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)=\frac{d}{du}\tan u\frac{d}{dx}{\sin }^{-1}\frac{3}{4}\\ ={\sec }^{2}u\left(\frac{1}{\sqrt{1-\frac{9}{16}}}\right)\\ =\frac{1}{\sqrt{7}}{\sec }^2\left(\frac{1}{2}{\sin }^{-1}\frac{3}{4}\right)\end{array}$