Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

$L$.$H$.$S$ $= 2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7}$
Using the formula $2 \tan ^{-1} x = \tan ^{-1} \frac{2x}{1-x^2}$,we get:
$= \tan ^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2} \right) + \tan ^{-1} \frac{1}{7}$
$= \tan ^{-1} \left( \frac{1}{1 - \frac{1}{4}} \right) + \tan ^{-1} \frac{1}{7}$
$= \tan ^{-1} \left( \frac{1}{\frac{3}{4}} \right) + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{1}{7}$
Now,using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
$= \tan ^{-1} \left( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \cdot \frac{1}{7}} \right)$
$= \tan ^{-1} \left( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right) = \tan ^{-1} \left( \frac{31}{17} \right) = R.H.S$.