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Mathematics Question on Inverse Trigonometric Functions

sin⁡[2cos⁡−1cot⁡(2tan⁡−112)]\sin \left[2 \cos^{-1} \cot \left(2 \tan^{-1} \frac{1}{2}\right) \right] is equal to

A

378\frac{3 \sqrt{7}}{8}

B

578\frac{5 \sqrt{7}}{8}

C

572\frac{5 \sqrt{7}}{2}

D

372\frac{3 \sqrt{7}}{2}

Answer

378\frac{3 \sqrt{7}}{8}

Explanation

Solution

sin⁡[2cos⁡−1cot⁡(2tan⁡−112)]\sin \left[2 \cos^{-1} \cot \left(2 \tan^{-1} \frac{1}{2}\right) \right] = \sin \left[2 \cos ^{-1} \cot \left\\{ \tan^{-1} \left(\frac{2\left(\frac{1}{2}\right)}{1- \frac{1}{4}}\right)\right\\} \right] = \sin \left[2 \cos ^{-1} \cot \left\\{ \tan ^{-1}\left(\frac{4}{3}\right) \right\\}\right] =sin⁡[2cos⁡−1cot⁡(cot⁡−1(34))]= \sin \left[2 \cos ^{-1} \cot \left( \cot^{-1} \left(\frac{3}{4}\right)\right) \right] =sin⁡[2cos⁡−1(34)]=sin⁡[cos⁡−1(2(916)−1)]= \sin \left[2 \cos ^{-1} \left(\frac{3}{4}\right) \right] = \sin \left[ \cos ^{-1} \left(2 \left(\frac{9}{16}\right) -1\right) \right] =sin⁡[cos⁡−1(98−1)]=sin⁡[cos⁡−1(18)]= \sin \left[ \cos ^{-1}\left(\frac{9}{8} -1\right) \right] = \sin \left[ \cos ^{-1} \left(\frac{1}{8}\right) \right] =sin⁡[sin⁡−1(1−164)]=\sin \left[\sin^{-1} \left(\sqrt{1- \frac{1}{64}}\right) \right] =6364=9×78=378 = \sqrt{\frac{63}{64}} = \frac{\sqrt{9 \times7}}{8}= \frac{3 \sqrt{ 7}}{8}