If ${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$, then

We have, ${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$, where $x \in [ - 1,1]$

$\Rightarrow$ $x < \cos \left( {{{\sin }^{ - 1}}x} \right)$

$\Rightarrow$ $x < \cos \left[ {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right]$ $\left[ {let{{\sin }^{ - 1}}x = \theta \Rightarrow \sin \theta = \frac{x}{1}} \right]$

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

$\Rightarrow$ ${x^2} < \frac{1}{2} \Rightarrow x < \pm \left( {\frac{1}{{\sqrt 2 }}} \right)$

……(i)
Also, $- 1 \le x \le 1$

…….(ii)
$therefore, - 1 \le x \le \frac{1}{{\sqrt 2 }}$

Alternate Method
$\frac{\pi }{2} - {\sin ^{ - 1}}x > {\sin ^{ - 1}}x$

$\frac{\pi }{2} > 2{\sin ^{ - 1}}x \Rightarrow \frac{\pi }{4} > {\sin ^{ - 1}}x$

$\frac{1}{{\sqrt 2 }} > x \Rightarrow \frac{1}{{\sqrt 2 }} < x \le 1$

We know that, ${\sin ^{ - 1}}x \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$