How do I figure out this trig question algebraically and not graphically?

Question

I am trying to solve:

$\tan(\sin^{-1}x)$

I think the answer is $\frac{x}{\sqrt{1-x^2}}$

But I was only able to solve that graphically. I created a triangle on the unit circle, and since the sin is x, then 1 side is x, the hypotenuse is 1, and the third side seems to be $\sqrt{1-x^2}$

But how does one solve this algebraically?

I get up to here:

$$\frac{\sin(\sin^{-1}x)}{\cos(\sin^{-1}x)}$$

Now what? The top is $x$, but what does the denominator resolve to? and why?

Any thoughts on how to do:

$$\sin(\tan^{-1}x)$$

Is the answer $\frac{x}{\sqrt{1+x^2}}$

Answer 1

If $\cos y>0$, note that $\tan(y)=\frac{\sin y}{\cos y}=\frac{\sin y}{\sqrt{1-\sin^2y}}$, so letting $x=\sin y$, $$\tan(\sin^{-1}x)=\frac x{\sqrt {1-x^2}}$$

The case for $\cos y<0$ can be solved similarly (take the negative square root).

Answer 2

Let, $\sin^{-1}x=y$, then $x=\sin(y)$. Now, $\tan(\sin^{-1}x)=\tan(y)=\frac{\sin(y)}{\cos(y)}=\frac{x}{\sqrt{1-x^2}}$.