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} \

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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 each term inside the bracket:$1$. $\sin ^{-1} \sqrt{\frac{2 - \sqrt{3}}{4}} = \sin ^{-1} \sqrt{\frac{4 - 2\sqrt{3}}{8}} = \sin ^{-1} \sqrt{\frac{(\sqrt{3}-1)^2}{8}} = \sin ^{-1} \left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)$.Since $\sin 15^{\circ} = \sin(45^{\circ}-30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$,we have $\sin ^{-1} \left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right) = 15^{\circ} = \frac{\pi}{12}$.$2$. $\cos ^{-1} \frac{\sqrt{12}}{4} = \cos ^{-1} \frac{2\sqrt{3}}{4} = \cos ^{-1} \frac{\sqrt{3}}{2} = 30^{\circ} = \frac{\pi}{6}$.$3$. $\sec ^{-1} \sqrt{2} = \cos ^{-1} \frac{1}{\sqrt{2}} = 45^{\circ} = \frac{\pi}{4}$.Now,substitute these values back into the expression:$E = \cos ^{-1} \left[ \cot \left( \frac{\pi}{12} + \frac{\pi}{6} + \frac{\pi}{4} \right) \right]$$E = \cos ^{-1} \left[ \cot \left( \frac{\pi + 2\pi + 3\pi}{12} \right) \right]$$E = \cos ^{-1} \left[ \cot \left( \frac{6\pi}{12} \right) \right] = \cos ^{-1} \left[ \cot \left( \frac{\pi}{2} \right) \right]$Since $\cot \frac{\pi}{2} = 0$,we have $E = \cos ^{-1}(0) = \frac{\pi}{2}$.

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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