Question

Prove that $2{\tan }^{-1}\frac{1}{5}+2{\tan }^{-1}\frac{1}{8}+{\sec }^{-1}\frac{5\sqrt{2}}{7}=\frac{\pi }{4}$

Answer 1

$2{\tan }^{-1}\left(\frac{1}{5}\right)+2{\tan }^{-1}\left(\frac{1}{8}\right){\sec }^{-2}\frac{5\sqrt{2}}{7}=\frac{\pi }{4}$

$2{\tan }^{-1}\left(\frac{1/5+1/8}{1-1/40}\right)+{\sec }^{-1}\left(\frac{5\sqrt{2}}{4}\right)=\frac{\pi }{4}$

$2{\tan }^{-1}\left(\frac{131}{393}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{4}$

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

${\tan }^{-1}\left(\frac{3}{393}\right)+{\tan }^{-2}\left(\frac{1}{7}\right)=\frac{\pi }{4}$

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

${\tan }^{-1}\left(\frac{21+4/28}{\frac{28-3}{28}}\right)={\tan }^{-1}\left(\frac{25}{25}\right)={\tan }^{-1}\left(1\right)$

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