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Solution

$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$ $2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ $\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$ $2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\sqrt{3}$ $\tan^{-1}\left[2\sin\left\{2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}\right]=\frac{\pi}{3}$

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Solution

Range of $\sin^{-1}x$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$ So A → IV Range of $\tan^{-1}x$ is $(-\frac{\pi}{2},\frac{\pi}{2})$ So B → II Range of $\csc^{-1}x$ is $[-\frac{\pi}{2},\frac{\pi}{2}] - {0}$ So C → I Range of $\sec^{-1}x$ is $[0,\pi] - {\frac{\pi}{2}}$ So D → III

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Solution