Step 1: Check candidates
For $x=-1$: $\tan^{-1}(-1) = -\frac\pi4$ and $\cot^{-1}(-1) = \frac{3\pi}{4}$. Sum of squares: $\frac{\pi^2}{16}+\frac{9\pi^2}{16} = \frac{10\pi^2}{16} = \frac{5\pi^2}{8}$.
Step 2: Others fail
For $x=1$: $\frac{\pi^2}{16}+\frac{\pi^2}{16} = \frac{\pi^2}{8}$. For $x=2$ and $x=-2$ the sums are different from $\frac{5\pi^2}{8}$ as the angles are not $\pm\frac\pi4$.
Step 3: Result
Option (A).
Final Answer:
x = -1.
\[ \boxed{\text{(A)}\ -1} \]