$2{\mathrm{Tan}}^{-1}\frac{1}{5}+{\mathrm{Sec}}^{-1}\frac{5\sqrt{2}}{7}+2{\mathrm{Tan}}^{-1}\frac{1}{8}=$

$$\begin{gathered} 2\left({\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7} \\ 2{\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{5}}+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{1-{\displaystyle \frac{1}{5}}.{\displaystyle \frac{1}{8}}}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7} \\ 2{\tan }^{-1}\left(\frac{1}{3}\right)+{\sec }^{-1}\frac{5\sqrt{2}}{7} \\ {\tan }^{-1}\left(\frac{2.{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}}\right)+{\cos }^{-1}\left(\frac{7}{5\sqrt{2}}\right) \\ {\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{7}\right) \\ {\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{3}{4}}.{\displaystyle \frac{1}{7}}}\right)={\tan }^{-1}\left(\frac{21+4}{28-3}\right)=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \end{gathered}$$