(2 ^ - 1 ( 1)/(5) + ^ - 1 ( 5)/(2) +2 ^ - 1 ( 1)/(8)) & is equal to:

Given,

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

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

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

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

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

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

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

$= {tantan}^{- 1}\dfrac{\dfrac{5}{4}}{\dfrac{5}{8}}$

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

$= 2$