If ${\tan }^{-1}\frac{1}{5}+\frac{1}{2}{\sec }^{-1}x+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{8},$then $x^2=$

$\Rightarrow {\tan }^{-1}\left[\frac{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle 5}}+\frac{{\displaystyle 1}}{{\displaystyle 8}}}}{{\displaystyle 1-\frac{{\displaystyle 1}}{{\displaystyle 5}}\times \frac{{\displaystyle 1}}{{\displaystyle 8}}}}\right]+\frac{1}{2}{\sec }^{-1}x=\frac{\pi }{8}$
$\begin{array}{l}\Rightarrow 2{\tan }^{-1}\frac{13}{39}+{\sec }^{-1}x=\frac{\pi }{4}\\ \Rightarrow 2{\tan }^{-1}\frac{1}{3}+{\sec }^{-1}x=\frac{\pi }{4}\\ \Rightarrow {\tan }^{-1}\frac{3}{1-\frac{1}{9}}+{\sec }^{-1}x=\frac{\pi }{4}\\ \Rightarrow {\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\sqrt{x^2-1}=\frac{\pi }{4}\\ \Rightarrow {\tan }^{-1}\left[\frac{\frac{3}{4}+\sqrt{x^2-1}}{1-\frac{3}{4}\sqrt{x^2-1}}\right]=\frac{\pi }{4}\end{array}$
$$\begin{gathered} \begin{array}{l}\Rightarrow 3+4\sqrt{x^2-1}=4-3\sqrt{x^2-1} \\ \Rightarrow 7\sqrt{x^2-1}=1\\ \Rightarrow \sqrt{x^2-1}=\frac{1}{7}\\ x^2-1=\frac{1}{49}\Rightarrow x^2=\frac{50}{49}\end{array} \end{gathered}$$