Mathematics - Point Question with Solution | TestHub

Given ${\sin }^{-1}x>{\cos }^{-1}x$ where $x\in (0,1)$ $$\begin{array}{rl}& \Rightarrow {\sin }^{-1}x>\frac{\pi }{2}-{\sin }^{-1}x\\ & \Rightarrow 2{\sin }^{-1}x>\frac{\pi }{2}\Rightarrow {\sin }^{-1}x>\frac{\pi }{4}\\ & \Rightarrow x>\sin \frac{\pi }{4}\Rightarrow x>\frac{1}{\sqrt{2}}\end{array}$$ Maximum value of ${\sin }^{-1}x$ is $\frac{\pi }{2}$ So, maximum value of $x$ is 1 . So, $x\in \left(\frac{1}{\sqrt{2}},1\right)$