#### Find the value of  $4 \tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{239}$

We have, $4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}$

$=2 \times \left ( 2 \tan^{-1}\frac{1}{5} \right )-\tan^{-1}\frac{1}{239}$

$=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{1-\left ( \frac{1}{5} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239}$    $\left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]$

$=2 \left [ \tan^{-1}\frac{\frac{2}{5}}{ \frac{24}{25} } \right ]-\tan^{-1}\frac{1}{239}$

$=2 \tan^{-1}\frac{5}{12}-\tan^{-1}\frac{1}{239}$

$=\left [ \tan^{-1}\frac{\frac{5}{6}}{1-\left (\frac{5}{12} \right )^{2}} \right ]-\tan^{-1}\frac{1}{239}$ $\left [ since, 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-\left ( x \right )^{2}} \right ]$

$=\tan^{-1}\frac{\frac{5}{6}}{1-\frac{25}{144}}-\tan^{-1}\frac{1}{139}$

$=\tan^{-1}\left ( \frac{144 \times 5}{119 \times 6} \right )-\tan^{-1}\frac{1}{239}$

$=\tan^{-1}\frac{120}{119}-\tan^{-1}\frac{1}{239}$

$=\tan^{-1}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}\left [ since, \tan^{-1} x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right ) \right ]$

$=\tan^{-1}\left [\frac{28680-119}{28441+120} \right ]$

$=\tan^{-1}\frac{28561}{28561}$

$=\tan^{-1}\left ( 1 \right )$

$=\tan^{-1}\left ( \tan \frac{\pi}{4} \right )$

$=\frac{\pi}{4}$

Hence, $4 tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}=\frac{\pi}{4}$