# The sum of possible values of

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

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

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

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

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

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 :-

$\frac{(x+1)+(x-1)}{1-\left(x^{2}-1\right)}=\frac{8}{31}$

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

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

$\Rightarrow \mathrm{x}=-8, \frac{1}{4}$

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

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

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

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

(Not possible)

Hence, $x=-8$