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

 For all $$|x| \geq 1$$, we have:

$$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$$

Setting up variables: Let $$\alpha = \sec^{-1}x$$. Then $$\csc^{-1}x = \frac{\pi}{2} - \alpha$$.

Range of $$\alpha$$: The principal value of $$\sec^{-1}x$$ lies in $$[0, \pi] \setminus \{\frac{\pi}{2}\}$$.

For $$x \geq 1$$: $$\alpha \in \left[0, \frac{\pi}{2}\right)$$

For $$x \leq -1$$: $$\alpha \in \left(\frac{\pi}{2}, \pi\right]$$

So the full domain is $$\alpha \in \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$$.

Expressing the function:

$$f(\alpha) = 16\left[\alpha^2 + \left(\frac{\pi}{2} - \alpha\right)^2\right]$$

$$= 16\left[\alpha^2 + \frac{\pi^2}{4} - \pi\alpha + \alpha^2\right]$$

$$= 16\left[2\alpha^2 - \pi\alpha + \frac{\pi^2}{4}\right]$$

$$= 32\alpha^2 - 16\pi\alpha + 4\pi^2$$

Finding the critical point:

$$f'(\alpha) = 64\alpha - 16\pi = 0$$

$$\alpha = \frac{\pi}{4}$$

Since $$f''(\alpha) = 64 > 0$$, this is a minimum. The point $$\alpha = \frac{\pi}{4}$$ lies in the domain (it is in $$[0, \frac{\pi}{2})$$).

Minimum value:$$f\left(\frac{\pi}{4}\right) = 32 \cdot \frac{\pi^2}{16} - 16\pi \cdot \frac{\pi}{4} + 4\pi^2$$

$$= 2\pi^2 - 4\pi^2 + 4\pi^2 = 2\pi^2$$

Maximum value: Since $$f(\alpha)$$ is a parabola opening upward with vertex at $$\alpha = \frac{\pi}{4}$$, the maximum on the domain occurs at the endpoint farthest from $$\frac{\pi}{4}$$.

At $$\alpha = 0$$: $$f(0) = 0 - 0 + 4\pi^2 = 4\pi^2$$

At $$\alpha = \pi$$: $$f(\pi) = 32\pi^2 - 16\pi^2 + 4\pi^2 = 20\pi^2$$

The maximum value is $$20\pi^2$$ (at $$\alpha = \pi$$, i.e., $$x = -1$$).

Sum of maximum and minimum $$= 20\pi^2 + 2\pi^2 = 22\pi^2$$

This matches Option B.