Question

Prove the followings
$2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$

Answer 1

Applying the trigonometric identities in the LHS of $2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$.

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

$={\tan }^{-1}\left(\frac{1}{1-\frac{1}{4}}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{1}{\frac{3}{4}}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{4}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\left(\frac{4}{3}\right)\left(\frac{1}{7}\right)}\right)$

$={\tan }^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right)$

$={\tan }^{-1}\left(\frac{31}{17}\right)$

$=RHS$