The maximum of ${\left(Se{c}^{-1}x\right)}^2+{\left(Co{\sec }^{-1}x\right)}^2$ is equal to

Let $I={\left({\sec }^{-1}x\right)}^2+{\left(\cos e{c}^{-1}x\right)}^2$

$I={(Se{c}^{-1}x+\cos e{c}^{-1}x)}^2-2{\sec }^{-1}x\left(\frac{\pi }{2}-{\sec }^{-1}x\right)$

$=\frac{\pi }{4}+2{\left(Se{c}^{-1}x\right)}^2-\pi {\sec }^{-1}x$

$=\frac{{\pi }^2}{4}+2\left[{\left({\sec }^{-1}x-\frac{\pi }{4}\right)}^2-\frac{{\pi }^2}{16}\right]$

$\begin{array}{l}=\frac{{\pi }^2}{4}-\frac{{\pi }^2}{8}+2\left[\frac{9{\pi }^2}{16}\right]\\ =\frac{{\pi }^2}{8}+\frac{9{\pi }^2}{8}=\frac{10{\pi }^2}{8}=\frac{5{\pi }^2}{4}\end{array}$