Prove $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$

To prove: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$

L.H.S. $=\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}$

$=\tan ^{-1} \frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11} \cdot \frac{7}{24}}$      $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1} \frac{\frac{48+77}{11 \times 24}}{\frac{11 \times 24-14}{11 \times 24}}$

$=\tan ^{-1} \frac{48+77}{264-14}=\tan ^{-1} \frac{125}{250}=\tan ^{-1} \frac{1}{2}=$ R.H.S.