1. Given equation

We need to solve

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

We will use the principal value conventions:

• $\tan^{-1} y \in \left(-\frac\pi2,\frac\pi2\right)$
• $\cot^{-1} y \in (0,\pi)$

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2. Rewrite the cot inverse term

Let

$$\theta=\cot^{-1}\left(\frac{1}{x-1}\right).$$

Then

$$\cot\theta=\frac{1}{x-1}.$$

So, when expressed in terms of $\tan^{-1}$ carefully by sign cases,

$$\cot^{-1}\left(\frac{1}{x-1}\right)= \begin{cases} \tan^{-1}(x-1), & x>1,\\[4pt] \pi+\tan^{-1}(x-1), & x<1. \end{cases}$$

(At $x=1$, the expression is undefined.)

Thus we solve in two cases.

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3. Case 1: $x>1$

Then the equation becomes

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

Using

$$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$$

with quadrant adjustment.

Here

$$a=x+1,\quad b=x-1.$$

So

$$a+b=2x,$$

and

$$ab=(x+1)(x-1)=x^2-1.$$

Hence

$$\tan\big(\tan^{-1}(x+1)+\tan^{-1}(x-1)\big)=\frac{2x}{1-(x^2-1)}=\frac{2x}{2-x^2}.$$

Therefore

$$\frac{2x}{2-x^2}=\frac{8}{31}.$$

Cross-multiplying,

$$62x=16-8x^2$$ $$8x^2+62x-16=0$$ $$4x^2+31x-8=0.$$

Solving:

$$4x^2+31x-8=(4x-1)(x+8)=0.$$

So

$$x=\frac14 \quad \text{or} \quad x=-8.$$

But this case requires $x>1$, so no solution from Case 1.

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4. Case 2: $x<1$

Then

$$\cot^{-1}\left(\frac{1}{x-1}\right)=\pi+\tan^{-1}(x-1).$$

So the equation becomes

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

This is impossible directly because left side is larger by $\pi$, so instead we interpret through tangent:

Let

$$A=\tan^{-1}(x+1),\qquad B=\cot^{-1}\left(\frac{1}{x-1}\right).$$

Then

$$A+B=\tan^{-1}\left(\frac{8}{31}\right).$$

Taking tangent on both sides,

$$\tan(A+B)=\frac{8}{31}.$$

Now

$$\tan A=x+1,$$

and from

$$\cot B=\frac{1}{x-1}$$

we get

$$\tan B=x-1.$$

Therefore

$$\tan(A+B)=\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}=\frac{2x}{2-x^2}.$$

So again,

$$\frac{2x}{2-x^2}=\frac{8}{31}.$$

This gives

$$4x^2+31x-8=0$$

with roots

$$x=\frac14,\,-8.$$

Now we must check the original equation and the domain $x<1$.

• $x=\frac14<1$ is allowed.
• $x=-8<1$ is allowed.

Check both in the original equation using principal values:

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5. Verification of roots

For $x=\frac14$:

$$\tan^{-1}\left(x+1\right)=\tan^{-1}\left(\frac54\right),$$ $$\cot^{-1}\left(\frac{1}{x-1}\right)=\cot^{-1}\left(\frac{1}{-3/4}\right)=\cot^{-1}\left(-\frac43\right).$$

Since principal value of $\cot^{-1}$ lies in $(0,\pi)$,

$$\cot^{-1}\left(-\frac43\right)=\pi-\cot^{-1}\left(\frac43\right)=\pi-\tan^{-1}\left(\frac34\right).$$

Thus

$$\tan^{-1}\left(\frac54\right)+\pi-\tan^{-1}\left(\frac34\right)$$

is clearly not equal to the acute angle $\tan^{-1}(8/31)$. So $x=\frac14$ is not a valid solution.

For $x=-8$:

$$\tan^{-1}(x+1)=\tan^{-1}(-7),$$ $$\cot^{-1}\left(\frac{1}{x-1}\right)=\cot^{-1}\left(-\frac19\right).$$

Now

$$\cot^{-1}\left(-\frac19\right)=\pi-\cot^{-1}\left(\frac19\right)=\pi-\tan^{-1}(9).$$

So LHS is

$$\tan^{-1}(-7)+\pi-\tan^{-1}(9).$$

Using angle relation,

$$\pi-\big(\tan^{-1}(7)+\tan^{-1}(9)\big).$$

Now

$$\tan^{-1}(7)+\tan^{-1}(9)=\pi+\tan^{-1}\left(\frac{7+9}{1-63}\right) =\pi+\tan^{-1}\left(-\frac{8}{31}\right) =\pi-\tan^{-1}\left(\frac{8}{31}\right).$$

Hence

$$\text{LHS}=\pi-\left(\pi-\tan^{-1}\left(\frac{8}{31}\right)\right) =\tan^{-1}\left(\frac{8}{31}\right).$$

So $x=-8$ is a valid solution.

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6. Sum of possible values of $x$

There is only one valid value:

$$x=-8.$$

So the sum is

$$-8=-\frac{32}{4}.$$

Hence the correct option is:

$$\boxed{\text{A}}$$

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7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.