How do I write a trig function that includes inverses in terms of another variable?

Question

It's been awhile since I've used trig and I feel stupid asking this question lol but here goes:

Given: $z = \tan(\arcsin(x))$

Question: How do I write something like that in terms of $x$?

Thanks! And sorry for my dumb question.

Answer 1

$$z = \tan[\arcsin(x)]$$

$$\arctan(z) = \arctan[\tan(\arcsin(x)] = \arcsin(x)$$

$$\sin[\arctan(z)] = \sin[\arcsin(x)] = x$$

Answer 2

Draw a triangle $ABC$, right-angled at $C$. Let us suppose that $\arcsin x=\angle A$.

Decide that the hypotenuse $AB$ is of length $1$ (it doesn't matter). Write $1$ next to the hypotenuse.

Then the side $BC$ opposite $\angle A$ has length $x$. The remaining side $AC$, by the Pythagorean Theorem, is $\sqrt{1-x^2}$. Write the appropriate lengths on the diagram.

Finally, read off $\tan A$ from the picture. It is $\frac{x}{\sqrt{1-x^2}}$.

This is not quite all there is to it. We need to check whether the result makes sense for all $x$. The only possible issues that can arise for angles not between $0$ and $\frac{\pi}{2}$ are issues of sign. Since $\arctan x$ is conventionally always between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and on this interval $\sin$ has the same sign as $\tan$, there is no problem.

If you want something like $\csc(\arctan x)$, do something similar. We could let the hypotenuse be $1$. But it may be easier to let the leg adjacent to $\angle A$ be $1$, and the leg opposite $\angle A$ be $x$. Then the hypotenuse is $\sqrt{1+x^2}$, and now you can read off the cosecant from the picture.