Question

Prove that : ${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{4}$

Answer 1

$ta{n}^{-1}\left(\frac{\left(\frac{1}{3}\right)+\left(\frac{1}{7}\right)}{1-\left(\frac{1}{21}\right)}\right)+ta{n}^{-1}\left(\frac{\left(\frac{1}{5}\right)+\left(\frac{1}{8}\right)}{1-\left(\frac{1}{40}\right)}\right)=ta{n}^{-1}\left(\frac{10}{20}\right)+ta{n}^{-1}\left(\frac{13}{39}\right)=ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{3}\right)=ta{n}^{-1}\left(\frac{\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)}{1-\left(\frac{1}{20}\right)}\right)ta{n}^{-1}\left(\frac{5}{5}\right)=ta{n}^{-1}1=\left(\frac{\pi }{4}\right)$