<s>[INST] Find the values of $x$ such that $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$. 
Find the values of $x$ such that 
  $$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$.

Checking for $x\in [-1,1]$
In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ comes out to be $2\tan^{-1}x$ hence the taken function comes out to be equal to $4\tan^{-1}x$ hence the function is clearly dependent on $x$.
Now checking for $x\in (1,\infty)$
In the taken domain $2\tan^{-1}x$ comes out to be $\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes independent of $x$. 
Now checking for $x\in (-\infty,-1)$
In the taken domain $2\tan^{-1}x$ comes out to be $-\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$  and hence the net sum becomes $-\pi$ therefore becomes, independent of $x$.
But the answer has been mentioned as just $x\in [1,\infty)$ Can anybody tell me why the second set has not been included.
 [/INST] Let $\tan^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2\iff-\pi\le2y\le\pi$
$$\implies\dfrac{2x}{1+x^2}=\sin2y$$
$$\sin^{-1}\dfrac{2x}{1+x^2}=\begin{cases}-2y-\pi &\mbox{if }2y<-\dfrac\pi2\iff x<-1 \\ 
2y & \mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\\ \pi-2y & \mbox{if }2y>\dfrac\pi2\end{cases}$$
Can you take it home from here?
See also: 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 $xy>1$
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