Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

We have,

${\sin }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}\Rightarrow 0<x<1$

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

$\Rightarrow {\tan }^{-1}x={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$

$\Rightarrow x^2=\sqrt{1-x^2}$

$\Rightarrow x^4+x^2-1=0\Rightarrow x^2=\frac{-1\pm \sqrt{5}}{2}$

as $x^2>0$

$\therefore x^2=\frac{\sqrt{5}-1}{2}$

$\Rightarrow 2x^2+1=\sqrt{5}$