Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\begin{array}{rl}& 16{\left({\sec }^{-1}x\right)}^2+{\left({\mathrm{cosec}}^{-1}x\right)}^2\\ & {\mathrm{Sec}}^{-1}x=a\in [0,\pi ]-\left\{\frac{\pi }{2}\right\}\\ & {\mathrm{cosec}}^{-1}x=\frac{\pi }{2}-a\\ & =16\left[a^2+{\left(\frac{\pi }{2}-a\right)}^2\right]=16\left[2a^2-\pi a+\frac{{\pi }^2}{4}\right]\\ & \max {]}_{a=\pi }=16\left[2{\pi }^2-{\pi }^2+\pi \frac{2}{4}\right]=20{\pi }^2\\ & \min {]}_{a=\frac{\pi }{4}}=16\left[\frac{2\times {\pi }^2}{16}-\frac{{\pi }^2}{4}+\frac{{\pi }^2}{4}\right]=2{\pi }^2\\ & \mathrm{Sum}=22{\pi }^2\end{array}$