Prove $\tan ^{-1}\left(\frac{2}{11}\right)+\tan ^{-1}\left(\frac{7}{24}\right)=\tan ^{-1}\left(\frac{1}{2}\right)$
We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
$L.H.S. = \tan ^{-1} \left( \frac{2}{11} \right) + \tan ^{-1} \left( \frac{7}{24} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{11} + \frac{7}{24}}{1 - (\frac{2}{11} \times \frac{7}{24})} \right)$
$= \tan ^{-1} \left( \frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} \right)$
$= \tan ^{-1} \left( \frac{125}{250} \right)$
$= \tan ^{-1} \left( \frac{1}{2} \right) = R.H.S.$
$L.H.S. = \tan ^{-1} \left( \frac{2}{11} \right) + \tan ^{-1} \left( \frac{7}{24} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{11} + \frac{7}{24}}{1 - (\frac{2}{11} \times \frac{7}{24})} \right)$
$= \tan ^{-1} \left( \frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} \right)$
$= \tan ^{-1} \left( \frac{125}{250} \right)$
$= \tan ^{-1} \left( \frac{1}{2} \right) = R.H.S.$
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