Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :

<p>Let $ f(x) = 16\left[\left(\sec^{-1} x\right)^2 + \left(\operatorname{cosec}^{-1} x\right)^2\right] $.</p> <p>We can express $ f(x) $ as:</p> <p>$ f(x) = 16\left[\left(\sec^{-1} x + \operatorname{cosec}^{-1} x\right)^2 - 2\left(\sec^{-1} x\right)\left(\frac{\pi}{2} - \sec^{-1} x\right)\right] $</p> <p>This simplifies to:</p> <p>$ f(x) = 16\left[\frac{\pi^2}{4} - \pi \sec^{-1} x + 2 \left(\sec^{-1} x\right)^2\right], \quad \text{where } \sec^{-1} x \in [0, \pi] - \left\{\frac{\pi}{2}\right\} $</p> <p>Further simplification gives:</p> <p>$ f(x) = 16\left[2\left(\sec^{-1} x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{4} - \frac{\pi^2}{8}\right] $</p> <p>For the maximum value when $ \sec^{-1} x = \pi $:</p> <p>$ \max = 16\left[2\pi^2 - \pi^2 + \frac{\pi^2}{4}\right] = 20\pi^2 $</p> <p>For the minimum value when $ \sec^{-1} x = \frac{\pi}{4} $:</p> <p>$ \min = 16\left[\frac{2 \times \pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4}\right] = 2\pi^2 $</p> <p>Therefore, the sum of the maximum and minimum values is:</p> <p>$ \text{Sum} = 22\pi^2 $</p>