$\text{tan⁡}\left(2{\tan }^{-1}\frac{1}{5}+{\sec }^{-1}\frac{\sqrt{5}}{2}+2{\tan }^{-1}\frac{1}{8}\right)$ is equal to

$\begin{array}{c}\tan \left(2\left({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right)\\ =\tan \left(2\left({\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{8}}}{1-\frac{1}{5}\times \frac{1}{8}}\right)\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right)\\ =\tan \left[2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right]\\ =\tan \left({\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{2}\right)\right)\\ =\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{4}\times \frac{1}{2}}\\ =2\end{array}$