Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of \(16\…

\(16\left({\left({\sec }^{-1}x\right)}^{2}+{\left({\csc }^{-1}x\right)}^{2}\right)............\text{(i)}\\ \text{we know that,}se{c}^{-1}x+\cos e{c}^{-1}x=\frac{\pi }{2}\\ \text{and}se{c}^{-1}x\in \left[0,\pi \right]-\left{\frac{\pi }{2}\right}\\ \text{now, put in (i), we get}\\ 16\left({\left({\sec }^{-1}x\right)}^{2}+{\left(\frac{\pi }{2}-se{c}^{-1}x\right)}^{2}\right)\\ letse{c}^{-1}x=y\\ 16\left(2{y}^{2}-\pi y+\frac{{\pi }^{2}}{4}\right)=0\\ \text{Now}\text{, }\text{it}\text{ }\text{is}\text{ }\text{maximum}\text{ }\text{at}y=\pi \\ \text{maximum}\text{ }\text{value}=16[2{\pi }^{2}-{\pi }^{2}+\frac{{\pi }^{2}}{4}]\\ =20{\pi }^{2}\\ \text{and}\text{ }\text{minimum}\text{ }\text{at}y=\frac{\pi }{4}\\ (y=-\frac{b}{2a}\text{for the quadratic in}y)\\ \text{minimum}\text{ }\text{value}=16[\frac{2\times {\pi }^{2}}{16}-\frac{{\pi }^{2}}{4}+\frac{{\pi }^{2}}{4}]\\ =2{\pi }^{2}\\ \text{And}\text{ }\text{sum}\text{ }\text{of}\text{ }\text{maximum}\text{ & }\text{minimum}\text{ value}=\\ 20{\pi }^{2}+2{\pi }^{2}=22{\pi }^{2}\)