Question

If $f\left(x\right)={\tan }^{-1}\left(\frac{1-\sqrt{x^2-1}}{\sqrt{x^2+2\sqrt{x^2-1}}}\right)+{\sin }^{-1}\left(\frac{\sqrt{x^2-1}}{|x|}\right)$ for $|x|\ge 1,$ then the number of solution(s) of the equation $\tan \left(f\right(x\left)\right)=|x^2-2|$ is

Answer 1

Put $x=\sec \theta$ where $\theta \in [0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ]$
$f\left(x\right)={\tan }^{-1}\left(\frac{1-|\tan \theta |}{\sqrt{{\sec }^2\theta +2|\tan \theta |}}\right)+{\sin }^{-1}\left(\frac{|\tan \theta |}{|\sec \theta |}\right)$

For $x\ge 1,$ $\theta \in [0,\frac{\pi }{2})$
$f\left(x\right)={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)+{\sin }^{-1}\left(\sin \theta \right)={\tan }^{-1}\tan (\frac{\pi }{4}-\theta )+\theta =\frac{\pi }{4}-\theta +\theta =\frac{\pi }{4}$

For $x\le -1,$ $\theta \in (\frac{\pi }{2},\pi ]$
$f\left(x\right)={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)+{\sin }^{-1}\left(\sin \theta \right)={\tan }^{-1}\left(\tan (\frac{\pi }{4}+\theta )\right)+{\sin }^{-1}\left(\sin \theta \right)$
We know that ${\tan }^{-1}\left(\tan x\right)=x-\pi$ for $x\in (\frac{\pi }{2},\frac{3\pi }{2})$
and ${\sin }^{-1}\left(\sin x\right)=\pi -x$ for $x\in (\frac{\pi }{2},\pi )$
$f\left(x\right)=\frac{\pi }{4}+\theta -\pi +(\pi -\theta )=\frac{\pi }{4}$

$\therefore \tan \left(f\right(x\left)\right)=1$ for $|x|\ge 1$

$|x^2-2|=1$
$\Rightarrow x^2-2=\pm 1$
$x^2=3,1$
$\Rightarrow x=\pm \sqrt{3},\pm 1$
Hence, there are four solutions.