<s>[INST] 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!
 [/INST] 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 $\arctan\dfrac{2x}{1-x^2}=y,\tan y=\dfrac{2x}{1-x^2},-\dfrac\pi2<y<\dfrac\pi2$
$\sec y=+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$
$\sin y=\dfrac{\tan y}{\sec y}=?$
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