Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given,

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

$=\tan \left(2\left({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right)$

$=\tan \left(2\left({\tan }^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-{\displaystyle \frac{1}{40}}}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right)$

$=\tan \left[2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]$

$=\tan \left[{\tan }^{-1}\left(\frac{2\times \frac{1}{3}}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]$

$=\tan \left[{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]$

$=\tan \left[{\tan }^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-{\displaystyle \frac{3}{8}}}\right]$

$={\mathrm{tantan}}^{-1}\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{5}{8}}}$

$={\mathrm{tantan}}^{-1}2$

$=2$