SOLUTION : L.HS. $=2 \tan ^{-1} \frac{1}{2} \div \tan ^{-1} \frac{1}{7} \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right]$
$=\tan ^{-1} \frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2} \div \tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{1}{1-\frac{1}{4}}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7} \\$
$=\tan ^{-1}\left[\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}\right] \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right] \\$
$=\tan ^{-1}\left[\frac{\frac{28-3}{21}}{\frac{21-4}{21}}\right]=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right)=\tan ^{-1} \frac{31}{17}=\text { R.H.S. }$
