Question

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

• A. $\frac{3\sqrt{7}}{8}$
• B. $\frac{5\sqrt{7}}{8}$
• C. $\frac{5\sqrt{7}}{2}$
• D. $\frac{3\sqrt{7}}{2}$

Answer 1

Answer: (A)

$\sin \left[2{\cos }^{-1}\cot \left(2{\tan }^{-1}\frac{1}{2}\right)\right]$
$=\sin \left[2{\cos }^{-1}\cot \left\{{\tan }^{-1}\left(\frac{2\left(\frac{1}{2}\right)}{1-\frac{1}{4}}\right)\right\}\right]$
$=\sin \left[2{\cos }^{-1}\cot \left\{{\tan }^{-1}\left(\frac{4}{3}\right)\right\}\right]$
$=\sin \left[2{\cos }^{-1}\cot \left({\cot }^{-1}\left(\frac{3}{4}\right)\right)\right]$
$=\sin \left[2{\cos }^{-1}\left(\frac{3}{4}\right)\right]=\sin \left[{\cos }^{-1}\left(2\left(\frac{9}{16}\right)-1\right)\right]$
$=\sin \left[{\cos }^{-1}\left(\frac{9}{8}-1\right)\right]=\sin \left[{\cos }^{-1}\left(\frac{1}{8}\right)\right]$
$=\sin \left[{\sin }^{-1}\left(\sqrt{1-\frac{1}{64}}\right)\right]$
$=\sqrt{\frac{63}{64}}=\frac{\sqrt{9\times 7}}{8}=\frac{3\sqrt{7}}{8}$