SOLUTION : L.H.S. $\tan ^{-1}\left(\frac{1}{2}\right) \div \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$
$=\tan ^{-}\left[\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2} \cdot \frac{1}{5}}\right]+\tan ^{-1}\left(\frac{1}{8}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right] \\$
$=\tan ^{-1}\left[\frac{\frac{5+2}{10}}{\left.\frac{10-1}{10}\right]}+\tan ^{-1} \frac{1}{8}=\tan ^{-1}\left[\frac{7}{10} \times \frac{10}{9}\right]+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{7}{9}+\tan ^{-1} \frac{1}{8}=\tan ^{-1}\left[\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{9} \cdot \frac{1}{8}}\right]\right. \\$
$=\tan ^{-1}\left[\frac{\frac{56+9}{72}}{\left.\frac{72-7}{72}\right]=\tan ^{-1}\left[\frac{65}{72} \times \frac{72}{65}\right]=\tan ^{-1}(1)=\frac{\pi}{4}}\right.$
