SOLUTION : $\quad \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \quad \text { L.H.S. }=\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8} \\$
$=\tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \cdot \frac{1}{7}}\right]+\tan ^{-1}\left[\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \cdot \frac{1}{8}}\right]=\tan ^{-1}\left[\frac{\frac{7+5}{35}}{1-\frac{1}{35}}\right]+\tan ^{-1}\left[\frac{\frac{8+3}{24}}{1-\frac{1}{24}}\right] \\$
$=\tan ^{-1}\left(\frac{12}{34}\right)+\tan ^{-1}\left(\frac{11}{23}\right)=\tan ^{-1}\left(\frac{6}{17}\right)+\tan ^{-1}\left(\frac{11}{23}\right)=\tan ^{-1}\left[\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \cdot \frac{11}{23}}\right] \\$
$=\tan ^{-1}\left[\frac{138+187}{17 \times 23} \frac{66}{17 \times 23}\right]=\tan ^{-1}\left[\frac{325}{325}\right]=\tan ^{-1}(1)=\frac{\pi}{4}=\text {R. H.S. }$
