Using the principal values of the inverse trigonometric functions the sum of the maximum and the minimum values of $16\left({\left(se{c}^{-1}x\right)}^2+{\left(\cos e{c}^{-1}x\right)}^2\right)$ is :

The given expression is $16\left({\left(se{c}^{-1}x\right)}^2+{\left(\cos e{c}^{-1}x\right)}^2\right)$

To get the minimum value of the above expression, substitute $x=\sqrt{2}$

$16\left({\left(\frac{\mathrm{\pi }}{4}\right)}^2+{\left(\frac{\mathrm{\pi }}{4}\right)}^2\right)=2{\mathrm{\pi }}^2$

To get the maximum value of the above expression substitute $x=-1$

$16\left({\left(\mathrm{\pi }\right)}^2+{\left(-\frac{\mathrm{\pi }}{2}\right)}^2\right)=20{\mathrm{\pi }}^2$

Hence, the required sum is $22{\mathrm{\pi }}^2$