Question

$2ta{n}^{-1}\left(\frac{1}{3}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{4}$

Answer 1

$2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

$={\tan }^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)$

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

$={\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\times \frac{1}{7}}\right)$ $[\because \frac{3}{4}\times \frac{1}{7}<1]$

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

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

$=\frac{\pi }{4}.$

Hence, proved.