if ${\sin }^{-1}x>{\cos }^{-1}x$ then x should lie in the interval:

Since, ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
$\begin{array}{l}\Rightarrow {\sin }^{-1}x>\frac{\pi }{2}-{\sin }^{-1}x\\ \Rightarrow {\sin }^{-1}x>\frac{\pi }{4}\\ \Rightarrow x>\sin \frac{\pi }{4}\\ \Rightarrow x>\frac{1}{\sqrt{2}}\end{array}$
Hence, x lies in the interval $\left(\frac{1}{\sqrt{2}},1\right)$.