If sin ^{-1} x

Question

(A) Given, $\sin ^{-1} x < \cos ^{-1} x$ Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$, we can write $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1}

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$-1 \leq x < \frac{1}{\sqrt{2}}$

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(A) Given, $\sin ^{-1} x Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$, we can write $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$. Substituting this into the inequality: $\frac{\pi}{2} - \cos ^{-1} x $\Rightarrow \frac{\pi}{2} $\Rightarrow \cos ^{-1} x > \frac{\pi}{4}$ Since $\cos 	heta$ is a strictly decreasing function in the interval $[0, \pi]$, applying $\cos$ on both sides reverses the inequality: $x $x Also, the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$. Combining $x $-1 \leq x < \frac{1}{\sqrt{2}}$

If $\sin ^{-1} x < \cos ^{-1} x$, then

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