The set of value sof x satisfying |sin^-1x|<|cos^-1x|, is

We know that $0 ≤ cos^{-1} x ≤ \pi $.

$ ∴ |cos^{-1}x|= cos^{-1} x$

It is evident from the graphs of $y = |sin^{-1}x|$ and $y = |cos^{-1}x|$ that

$|sin^{-1}x| < |cos^{-1}x|$ for all x ∈$ [-1, 1/\sqrt{2})$