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

$$\begin{gathered} 2\left[{\tan }^{-1} \frac{1}{5}+{\tan }^{-1} \frac{1}{8}\right]+{\tan }^{-1} \frac{1}{7} \\ =2{\tan }^{-1} \left[\frac{1}{3}\right]+{\tan }^{-1} \frac{1}{7}={\tan }^{-1} \left(\frac{3}{4}\right)+{\tan }^{-1} \left(\frac{1}{7}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{7}}}{1-\frac{3}{4}.\frac{1}{7}}\right)={\tan }^{-1} 1=\frac{\pi }{4} \end{gathered}$$