Question

Prove the followings.${\tan }^{-1}\left(\frac{2}{11}\right)+{\tan }^{-1}\left(\frac{7}{24}\right)={\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

Applying the trigonometric identities in the LHS of ${\tan }^{-1}\left(\frac{2}{11}\right)+{\tan }^{-1}\left(\frac{7}{24}\right)={\tan }^{-1}\left(\frac{1}{2}\right)$

${\tan }^{-1}\left(\frac{2}{11}\right)+{\tan }^{-1}\left(\frac{7}{24}\right)={\tan }^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\times \frac{7}{24}}\right)$

$={\tan }^{-1}\left(\frac{\frac{2\left(24\right)+7\left(11\right)}{\left(11\right)\left(24\right)}}{\frac{\left(11\right)\left(24\right)-14}{\left(11\right)\left(24\right)}}\right)$

$={\tan }^{-1}\left(\frac{48+77}{264-14}\right)$

$={\tan }^{-1}\left(\frac{125}{250}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\right)$

$=RHS$