Question

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

Answer 1

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

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\times \frac{1}{8}}\right)$

On application of ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

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

$={\tan }^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times \frac{11}{23}}\right)={\tan }^{-1}\left(\frac{138+187}{391-66}\right)$

$={\tan }^{-1}\left(\frac{325}{325}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$

hence proved