How do I write a trig function that includes inverses in terms of another variable? 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.
 A: $$z = \tan[\arcsin(x)]$$
$$\arctan(z) = \arctan[\tan(\arcsin(x)] = \arcsin(x)$$
$$\sin[\arctan(z)] = \sin[\arcsin(x)] = x$$
A: 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.
