The value of cos ^{-1} left[ cot left{ sin ^{ | vedclass

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(D) Let the expression be $E = \cos ^{-1} \left[ \cot \left\{ \sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} + \cos ^{-1} \frac{\sqrt{12}}{4} + \sec ^{-1} \sq

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$\frac{\pi}{2}$

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(D) Let the expression be $E = \cos ^{-1} \left[ \cot \left\{ \sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} + \cos ^{-1} \frac{\sqrt{12}}{4} + \sec ^{-1} \sqrt{2} \right\} \right]$.First,simplify the terms inside the bracket:$1$. $\sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} = \sin ^{-1} \sqrt{\frac{4-2\sqrt{3}}{8}} = \sin ^{-1} \sqrt{\frac{(\sqrt{3}-1)^2}{8}} = \sin ^{-1} \left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right) = \frac{\pi}{12} = 15^{\circ}$.$2$. $\cos ^{-1} \frac{\sqrt{12}}{4} = \cos ^{-1} \frac{2\sqrt{3}}{4} = \cos ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6} = 30^{\circ}$.$3$. $\sec ^{-1} \sqrt{2} = \cos ^{-1} \frac{1}{\sqrt{2}} = \frac{\pi}{4} = 45^{\circ}$.Summing these angles: $\frac{\pi}{12} + \frac{\pi}{6} + \frac{\pi}{4} = \frac{\pi + 2\pi + 3\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$.Now,substitute back into the expression:$E = \cos ^{-1} \left( \cot \frac{\pi}{2} \right) = \cos ^{-1} (0) = \frac{\pi}{2}$.

The value of $\cos ^{-1} \left[ \cot \left\{ \sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}} + \cos ^{-1} \frac{\sqrt{12}}{4} + \sec ^{-1} \sqrt{2} \right\} \right]$ is

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