Inverse trigonometry: How to find $x$ When $\sin^{-1}\left(\frac{2x}{1+x²}\right)=2\tan^{-1}(x)$? I thought that it was for all $x\in\mathbb{R}$ but it was incorrect, so please help!
 A: Use 
Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
OR
Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?
$$2\arctan x=\begin{cases} \arctan\dfrac{2x}{1-x^2} &\mbox{if } x^2<1\\ \pi+\arctan\dfrac{2x}{1-x^2} & \mbox{if } x^2>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x^2=1\end{cases} $$
Finally if $\arctan\dfrac{2x}{1-x^2}=y,\tan y=\dfrac{2x}{1-x^2},-\dfrac\pi2<y<\dfrac\pi2$
$\sec y=+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$
$\sin y=\dfrac{\tan y}{\sec y}=?$
A: Let $\tan^{-1}x=y,-\dfrac\pi2<y<\dfrac\pi2\ \ \ \ (1)$
and $\sin^{-1}\dfrac{2x}{1+x^2}=\sin^{-1}(\sin2y)=n\pi+(-1)^n(2y)=f(y)$  where $n$ is an integer such that $-\dfrac\pi2\le f(y)\le\dfrac\pi2$
If $n=0,-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1$
If $n=1,-\dfrac\pi2\le\pi-2y\le\dfrac\pi2\iff\dfrac{3\pi}2\ge2y\ge\dfrac\pi2$  but $2y<\pi$ by $(1)$ $\implies\dfrac\pi2\le2y<\pi\implies x\ge1$
Similarly,  $n=-1,-\dfrac\pi2\le-\pi-2y\le\dfrac\pi2\iff x\le1$
A: Let $f(x)=\sin^{-1}\left(\frac{2x}{1+x²}\right)-2\tan^{-1}(x)$ and examine 
$$f’(x) =\left( \frac{1-x^2}{|1-x^2|}-1\right)\frac2{1+x^2}$$
which is apparent that, only for $x\in[-1,1]$, $f’(x)=0$ with the two functions equal to each other. 
