Let x in (0, 1). The set of all x such that s | vedclass

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(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this int

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$\left( \frac{1}{\sqrt{2}}, 1 \right)$

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(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this into the inequality,we get:$\sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x$$2 \sin^{-1} x > \frac{\pi}{2}$$\sin^{-1} x > \frac{\pi}{4}$Since the sine function is strictly increasing on the interval $[0, 1]$,we apply the sine function to both sides:$x > \sin\left( \frac{\pi}{4} \right)$$x > \frac{1}{\sqrt{2}}$Given the domain $x \in (0, 1)$,the solution set is $x \in \left( \frac{1}{\sqrt{2}}, 1 \right)$.

Let $x \in (0, 1)$. The set of all $x$ such that $\sin^{-1} x > \cos^{-1} x$ is the interval

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