Inverse Trigonometric Functions

Let $a=\sec^{-1}x$ and $b=\csc^{-1}x$.Using principal ranges, for $x\ge1$, $a\in[0,\frac{\pi}{2})$ and $b\in(0,\frac{\pi}{2}]$.For $x\le-1$, $a\in(\frac{\pi}{2},\pi]$ and $b\in[-\frac{\pi}{2},0)$.Since $\sec a=x$ and $\csc b=x$, we have $\cos a=\sin b$.With these principal ranges, this gives $a+b=\frac{\pi}{2}$, so $b=\frac{\pi}{2}-a$.Thus the expression is $E=16(a^2+b^2)=16\left[a^2+\left(\frac{\pi}{2}-a\right)^2\right]$.The quadratic has vertex at $a=\frac{\pi}{4}$.For $x\ge1$, $a\in[0,\frac{\pi}{2})$, so the minimum occurs at $a=\frac{\pi}{4}$.This gives $E_{\min}=16\left(\frac{\pi^2}{16}+\frac{\pi^2}{16}\right)=2\pi^2$.The maximum in this branch is $4\pi^2$.For $x\le-1$, $a\in(\frac{\pi}{2},\pi]$, so $E$ increases as $a$ increases.The maximum occurs at $a=\pi$, where $b=-\frac{\pi}{2}$.Thus $E_{\max}=16\left(\pi^2+\frac{\pi^2}{4}\right)=20\pi^2$.The limiting value in this branch is $4\pi^2$, which is larger than the global minimum $2\pi^2$.Therefore the overall minimum is $2\pi^2$ and the overall maximum is $20\pi^2$.Their sum is $2\pi^2+20\pi^2=22\pi^2$.Hence the correct option is $22\pi^2$, i.e. option D.