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

Given, ${\sin }^{-1}\mathrm{x}<{\cos }^{-1}\mathrm{x}$

$\Rightarrow {\cos }^{-1}\mathrm{x}>\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\mathrm{x}$      $\left({\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right)$

$\Rightarrow 2{\cos }^{-1}x>\frac{\mathrm{\pi }}{2}$

$\Rightarrow {\cos }^{-1}x>\frac{\mathrm{\pi }}{4}$

$\Rightarrow x<\cos \left(\frac{\mathrm{\pi }}{4}\right)$         (As $\cos \theta$ is decreasing function.)

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

Domain of  ${\cos }^{-1}\theta$ is $\left[-1,1\right] \dots \dots \left(2\right)$

Combining equation (1) and (2) we get,

$x\in [-1,\frac{1}{\sqrt{2}})$