Question

Show that $\tan {}^{-1}\left(\frac{1}{5}\right)+\tan {}^{-1}\left(\frac{1}{7}\right)+\tan {}^{-1}\left(\frac{1}{3}\right)+\tan {}^{-1}\left(\frac{1}{8}\right)=\frac{\pi }{4}$.

Answer 1

${\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}\times \frac{1}{8}}\right)$

$={\tan }^{-1}\left(\frac{12}{34}\right)+{\tan }^{-1}\left(\frac{11}{23}\right)$

$={\tan }^{-1}\left(\frac{\frac{12}{34}+\frac{11}{23}}{1-\frac{12}{34}\times \frac{11}{23}}\right)$

$={\tan }^{-1}\left(\frac{650}{650}\right)$

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

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