Let $x \in(0,1)$. The set of all $x$ such that $\sin ^{-1} x>\cos ^{-1} x$, is the interval:
Let $x \in(0,1)$. The set of all $x$ such that $\sin ^{-1} x>\cos ^{-1} x$, is the interval:
$\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$
$\left(\frac{1}{\sqrt{2}}, 1\right)$
$(0,1)$
$\left(0, \frac{\sqrt{3}}{2}\right)$
Solution
Given $\sin ^{-1} x>\cos ^{-1} x$ where $x \in(0,1)$
$
\begin{aligned}
&\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{aligned}
$
Maximum value of $\sin ^{-1} x$ is $\frac{\pi}{2}$
So, maximum value of $x$ is 1 . So, $x \in\left(\frac{1}{\sqrt{2}}, 1\right)$