Let's say we have this-
$$\arctan(\frac{2x}{1-x^2})$$
Now this equals-
$$2\arctan x, x\in[-1,1]$$ $$-\pi+2\arctan x, x\in[1,\infty]$$ $$\pi+2\arctan x, x\in[-\infty,-1]$$
Is there an easy way to find out the definition of such inverse trig functions. I mean what is the way to find out for what values of $x$ which formula to use.(How should I derive the above ranges of $x$ for which I should add $\pi$ for eg in the above example). I could somehow manage to do it by some complicated method with lots of steps that seems too hectic. How to I do it in an easy way.
In general,how should I deal with any inverse trigonometric function such as this or $\arctan\frac{3x-x^3}{1-3x^2}$ etc. and easily find their definition for various values of $x$.
Thanks for any help!
If $x\in\left(-1,1\right)$, then$$\tan\bigl(2\arctan(x)\bigr)=\frac{2\tan(\arctan x)}{1-\tan^2(\arctan x)}=\frac{2x}{1-x^2}.\tag1$$Since $2\arctan(x)\in\left(-\frac\pi2,\frac\pi2\right)$, it follows from this that$$2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right).$$
What if $x>1$? Then $(1)$ still holds, but now $2\arctan x\in\left(\frac\pi2,\frac{3\pi}2\right)$. But $2\arctan(x)-\pi\in\left(-\frac\pi2,\frac\pi2\right)$ and, since $\tan$ is periodc with period $\pi$, it follows that$$2\arctan(x)-\pi=\arctan\left(\frac{2x}{1-x^2}\right).$$
The other case is similar.