If $2\tan^{-1}x + \sin^{-1} \frac{2x}{1+x^2}$, find the values for $x$ for which the function is independent of $x$

Question

For $x>1$, then $$2\tan^{-1} x = \pi -2\sin^{-1} \frac{2x}{1+x^2}$$

For $x<-1$

$$2\tan^{-1} x =-\pi -2\sin^{-1} \frac{2x}{1+x^2}$$

So $x\in (-\infty, -1) \cup (1, \infty)$

But the given answer is only $(1,\infty)$

Is there any specific reason why we don’t consider the left part, or is it just a mistake?

Answer 1

Your answer is correct! It's vividly seen in the graph below: