Question

The value of : ${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{8}-\frac{\pi }{4}$

Answer 1

L.H.S. $={\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}{\tan }^{-1}\frac{1}{8}$
$={\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}\times \frac{1}{8}}\right)$
$={\tan }^{-1}\left(\frac{7+5}{35-1}\right)+{\tan }^{-1}\left(\frac{8+3}{24-1}\right)$
$={\tan }^{-1}\frac{12}{24}+{\tan }^{-1}\frac{11}{23}$
$={\tan }^{-1}\frac{6}{17}+{\tan }^{-1}\frac{11}{23}$
$={\tan }^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times \frac{11}{23}}\right)$
$={\tan }^{-1}\left(\frac{138+187}{391-66}\right)$
$={\tan }^{-1}\left(\frac{325}{325}\right)={\tan }^{-1}1$
$=\frac{\pi }{4}=$ R.H.S.