Question

Prove that
${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{6}{17}$

Answer 1

${\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}.\frac{1}{7}}\right)$

$={\tan }^{-1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right)$

$={\tan }^{-1}\left(\frac{12}{34}\right)$

$={\tan }^{-1}\left(\frac{6}{17}\right)$