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

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$- \frac{33}{4}$

$- \frac{31}{4}$

$- \frac{32}{4}$

$- \frac{30}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} The given eqution is \tan^{- 1} (x + 1) + \cot^{- 1} (\frac{1}{x - 1}) = \tan^{- 1} (\frac{8}{31}) \\ it implies \tan^{- 1} (x + 1) + \tan^{- 1} (x - 1) = \tan^{- 1} (\frac{8}{31}) \\ Hence, \Rightarrow \tan^{- 1} (\frac{x + 1 + x - 1}{1 - (x + 1) (x - 1)}) = \tan^{- 1} (\frac{8}{31}) \\ \Rightarrow \frac{2 x}{2 - x^{2}} = \frac{8}{31} \\ \Rightarrow 31 x = 8 - 4 x^{2} \\ \Rightarrow 4 x^{2} + 31 x - 8 = 0 \\ x = - 8, \frac{1}{4} \\ by substituting the values of x in the given equation, conclude that x = - 8 only the solution \end{array}$