The sum of possible values


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 :

  1. $-\frac{32}{4}$

  2. $-\frac{31}{4}$

  3. $-\frac{30}{4}$

  4. $-\frac{33}{4}$

Correct Option: 1


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

Taking tangent both sides :-


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

$\Rightarrow 4 x^{2}+31 x-8=0$

But, if $x=\frac{1}{4}$

$\tan ^{-1}(x+1) \in\left(0, \frac{\pi}{2}\right)$

$\& \cot ^{-1}\left(\frac{1}{x-1}\right) \in\left(\frac{\pi}{2}, \pi\right)$

$\Rightarrow \mathrm{LHS}>\frac{\pi}{2} \& \mathrm{RHS}<\frac{\pi}{2}$

(Not possible)

Hence, $x=-8$

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