Question

$\tan \left[3{\tan }^{-1}\left(\frac{1}{5}\right),-\frac{\pi }{4}\right]$ is equal to

• A. $-\frac{13}{46}$
• B. $-\frac{11}{46}$
• C. $-\frac{7}{46}$
• D. $-\frac{4}{23}$
• E. $-\frac{9}{46}$

Answer 1

Answer: (E)

$\tan \left\{3{\tan }^{-1}\left(\frac{1}{5}\right)-\frac{\pi }{4}\right\}$
$=\tan \left\{2{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)-{\tan }^{-1}(1)\right\}$
$=\tan \left\{{\tan }^{-1}\left(\frac{2\times \frac{1}{5}}{1-\frac{1}{25}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{5}-1}{1+\frac{1}{5}}\right)\right\}$
$=\tan \left\{{\tan }^{-1}\left(\frac{2}{5}\times \frac{25}{24}\right)+{\tan }^{-1}\left(-\frac{4}{5}\times \frac{5}{6}\right)\right\}$
$=\tan \left\{{\tan }^{-1}\left(\frac{5}{12}\right)-{\tan }^{-1}\left(\frac{2}{3}\right)\right\}$
$=\tan \left({\tan }^{-1}\frac{\frac{5}{12}-\frac{2}{3}}{1+\frac{5}{18}}\right)$
$=\tan \left\{{\tan }^{-1}\left(\frac{-3}{12}\times \frac{18}{23}\right)\right\}$
$=-\frac{9}{46}$