Question

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

Answer 1

$2ta{n}^{-1}\left(\frac{1}{5}\right)+se{c}^{-1}\left(\frac{5\sqrt{2}}{7}\right)+2ta{n}^{-1}\left(\frac{1}{8}\right)$

$=2\left[ta{n}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{40}}\right)\right]+se{c}^{-1}\left(\frac{5\sqrt{2}}{7}\right)$

$=2\left[ta{n}^{-1}\left(\frac{1}{3}\right)\right]+se{c}^{-1}\left(\frac{5\sqrt{2}}{7}\right)$

$=ta{n}^{-1}\frac{3}{4}+se{c}^{-1}\left(\frac{5\sqrt{2}}{7}\right)$

$=ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{7}$

$=ta{n}^{-1}\left(1\right)=\frac{\pi }{4}$