If $\cos ^{-1} x > \sin ^{-1} x$, then x lies in the interval

Question

$\begin{array}{1 1} \bigg(\large\frac{1}{2},1 \bigg) \\ (0,1) \\ \bigg[-1, \frac{1}{\sqrt {2}}\bigg] \\ [-1,1] \end{array}$