The sum of possible values of $$x$$ for $$\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\left(\frac{8}{31}\right)$$ is:

We need to solve $$\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\left(\frac{8}{31}\right)$$.

First, we convert $$\cot^{-1}\left(\frac{1}{x-1}\right)$$. For $$x - 1 > 0$$: $$\cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}(x-1)$$. For $$x - 1 < 0$$: $$\cot^{-1}\left(\frac{1}{x-1}\right) = \pi + \tan^{-1}(x-1)$$.

Case 1: $$x > 1$$. The equation becomes $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31}$$.

Using the addition formula $$\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}$$ (when $$AB < 1$$): $$\tan^{-1}\frac{(x+1)+(x-1)}{1-(x+1)(x-1)} = \tan^{-1}\frac{2x}{1-(x^2-1)} = \tan^{-1}\frac{2x}{2-x^2}$$.

Setting this equal to $$\tan^{-1}\frac{8}{31}$$: $$\frac{2x}{2-x^2} = \frac{8}{31}$$, giving $$62x = 16 - 8x^2$$, so $$8x^2 + 62x - 16 = 0$$, i.e., $$4x^2 + 31x - 8 = 0$$.

Solving: $$x = \frac{-31 \pm \sqrt{961 + 128}}{8} = \frac{-31 \pm \sqrt{1089}}{8} = \frac{-31 \pm 33}{8}$$.

So $$x = \frac{2}{8} = \frac{1}{4}$$ or $$x = \frac{-64}{8} = -8$$. Since we assumed $$x > 1$$, neither value works in this case.

Case 2: $$x < 1$$ (and $$x \neq 1$$). The equation becomes $$\tan^{-1}(x+1) + \pi + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31}$$.

So $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} - \pi$$.

Now if $$(x+1)(x-1) = x^2 - 1 > 1$$, i.e., $$x^2 > 2$$, then $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{2x}{2-x^2} - \pi$$ (when $$x < 0$$ and the product exceeds 1).

For $$x < 0$$ with $$|x| > \sqrt{2}$$: $$\tan^{-1}\frac{2x}{2-x^2} - \pi = \tan^{-1}\frac{8}{31} - \pi$$, so $$\frac{2x}{2-x^2} = \frac{8}{31}$$. This gives the same quadratic $$4x^2 + 31x - 8 = 0$$, so $$x = -8$$ (valid since $$-8 < 1$$ and $$|-8| > \sqrt{2}$$).

For $$x^2 < 2$$ (i.e., $$-\sqrt{2} < x < 1$$): $$\tan^{-1}\frac{2x}{2-x^2} = \tan^{-1}\frac{8}{31} - \pi$$, giving $$\frac{2x}{2-x^2} = \frac{8}{31}$$ again, so $$x = \frac{1}{4}$$ (valid since $$\frac{1}{4} < 1$$). Let us verify: $$\tan^{-1}(5/4) + \cot^{-1}\frac{1}{-3/4}$$. Here $$\frac{1}{x-1} = \frac{1}{-3/4} = -\frac{4}{3}$$. $$\cot^{-1}(-4/3) = \pi - \cot^{-1}(4/3) = \pi - \tan^{-1}(3/4)$$. So the LHS = $$\tan^{-1}(5/4) + \pi - \tan^{-1}(3/4) = \pi + \tan^{-1}\frac{5/4 - 3/4}{1 + 15/16} = \pi + \tan^{-1}\frac{1/2}{31/16} = \pi + \tan^{-1}\frac{8}{31}$$. This gives $$\pi + \tan^{-1}\frac{8}{31} \neq \tan^{-1}\frac{8}{31}$$. So $$x = \frac{1}{4}$$ does not satisfy the original equation.

Let us recheck with $$x = -8$$: $$\tan^{-1}(-7) + \cot^{-1}\frac{1}{-9} = \tan^{-1}(-7) + \cot^{-1}(-1/9)$$. Now $$\cot^{-1}(-1/9) = \pi - \cot^{-1}(1/9) = \pi - \tan^{-1}(9)$$. So LHS = $$\tan^{-1}(-7) + \pi - \tan^{-1}(9) = \pi - \tan^{-1}(7) - \tan^{-1}(9)$$. Using the formula: $$\tan^{-1}(7) + \tan^{-1}(9) = \pi + \tan^{-1}\frac{16}{1-63} = \pi + \tan^{-1}\frac{16}{-62} = \pi - \tan^{-1}\frac{8}{31}$$. So LHS = $$\pi - (\pi - \tan^{-1}\frac{8}{31}) = \tan^{-1}\frac{8}{31}$$. This works!

So the only valid solution is $$x = -8$$, and the sum of possible values is $$-8 = -\frac{32}{4}$$.

This matches Option A: $$-\frac{32}{4}$$.