\[\text { Prove that: } 2 \tan ^{-1}\left(\frac{1}{5}\right)+\sec ^{-1}\left(\frac{5 \sqrt{2}}{7}\right)+2 \tan ^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}\]

    \[\begin{array}{l} 2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8} \\ =2\left[\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right]+\sec ^{-1} \frac{5 \sqrt{2}}{7} \\ \left.=2 \tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \times \frac{1}{8}}\right\}+\tan ^{-1} \sqrt{\frac{5 \sqrt{2}}{7}}\right]^{2}-1\left[\because \sec ^{-1} x=\tan ^{-1} \sqrt{x^{2}-1}\right] \\ =2 \tan ^{-1} \frac{13}{39}+\tan ^{-1} \frac{1}{7} \\ =2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7} \\ =\tan ^{-1}\left[\frac{2 \times 1 / 3}{1-(1 / 3)^{2}}\right]+\tan ^{-1} \frac{1}{7}\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}, \text { if }|x|<1\right] \\ =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1}\left[\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \times \frac{1}{7}}\right]=\tan ^{-1} 1=\frac{\pi}{4} \end{array}\]