Question

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

The correct option is E $-\frac{9}{46}$

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