If $\begin{vmatrix}cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\end{vmatrix} < \frac{\pi}{3}, $ then x belongs to the interval

$(-\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}})$

We have,

$\begin{vmatrix}cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\end{vmatrix} < \frac{\pi}{3}$

$⇒ 0 < cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)< \frac{\pi}{3}$     $[∵ cos^{-1} x ∈ [0, \pi ]]$

$⇒ \frac{1}{2}< \frac{1-x^2}{1+x^2} < 1 $

$⇒ 1 + x^2 < 2 - 2x^2 < 2 + 2x^2 $

$⇒ 1 + x^2 < 2-2x^2 $ and $ 2 - 2x^2 < 2 +2x^2 $

$⇒ 3x^2 - 1 < 0 $ and $ 4x^2 > 0 ⇒ -\frac{1}{\sqrt{3}} < x< \frac{1}{\sqrt{3}}$