Simplifying Inverse Trig Function

Question

I'm trying to figure out how to simplify this expression but I'm not quite sure on how to approach this question. How should I approach this question? Any help is greatly appreciated!

$\tan(\sin^{-1}(\frac{y}{5})) $

Answer 1

You are looking for the tangent of the angle $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ whose sine is $\frac{y}{5}$. Draw a right triangle in quadrant I (or IV) with base angle at the origin and label the vertical side $y$ and the hypotenuse $5$. Label the side on the $x$-axis as $\sqrt{25-y^2}$. Then you should see that $\tan\theta=\dfrac{y}{\sqrt{25-y^2}}$.

Answer 2

If $\frac{y}{5} \in [-1,1]$, we have :

$$\tan(\sin^{-1}(\frac{y}{5}))=\frac {\sin(\sin^{-1}(\frac{y}{5}))}{\cos(\sin^{-1}(\frac{y}{5}))}=\frac{\frac{y}{5}}{\sqrt{1-(\frac{y}{5}})^2}$$

Because $\forall x \in [-1,1]$ we have $\cos(\arcsin(x))=\sin(\arccos(x))=\sqrt{1-x^2}$.