Question:
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 :
Correct Option: 1
Solution:
$\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1} \frac{8}{31}$
Taking tangent both sides :-
$\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}$
$\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}$
$\Rightarrow 4 x^{2}+31 x-8=0$
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, $x=-8$