The solution set of the inequation cos ^{-1} | vedclass

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(D) We are given the inequation $\cos ^{-1} x < \sin ^{-1} x$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}

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

Vedclass · Answer

(D) We are given the inequation $\cos ^{-1} x We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.Substituting this into the inequation,we get:$\frac{\pi}{2} - \sin ^{-1} x $\frac{\pi}{2} $\sin ^{-1} x > \frac{\pi}{4}$Taking the sine of both sides (since $\sin x$ is an increasing function in its domain):$x > \sin\left(\frac{\pi}{4}\right)$$x > \frac{1}{\sqrt{2}}$Since the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$,we must have $x \le 1$.Therefore,the solution set is $x \in \left(\frac{1}{\sqrt{2}}, 1\right]$.

The solution set of the inequation $\cos ^{-1} x < \sin ^{-1} x$ is

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