Prove $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

(N/A) $L$.$H$.$S$ = $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}$
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$,we group the terms:
$= \left( \tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{7} \right) + \left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{8} \right)$
$= \tan ^{-1} \left( \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}} \right) + \tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}} \right)$
$= \tan ^{-1} \left( \frac{\frac{12}{35}}{\frac{34}{35}} \right) + \tan ^{-1} \left( \frac{\frac{11}{24}}{\frac{23}{24}} \right)$
$= \tan ^{-1} \left( \frac{12}{34} \right) + \tan ^{-1} \left( \frac{11}{23} \right) = \tan ^{-1} \left( \frac{6}{17} \right) + \tan ^{-1} \left( \frac{11}{23} \right)$
$= \tan ^{-1} \left( \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}} \right) = \tan ^{-1} \left( \frac{\frac{138 + 187}{391}}{\frac{391 - 66}{391}} \right)$
$= \tan ^{-1} \left( \frac{325}{325} \right) = \tan ^{-1} (1) = \frac{\pi}{4} = R.H.S$