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

• A. $-\frac{32}{4}$
• B. $-\frac{31}{4}$
• C. $-\frac{30}{4}$
• D. $-\frac{33}{4}$

Answer 1

Answer: (A)

${\tan }^{-1}(x+1)+{\cot }^{-1}\left(\frac{1}{x-1}\right)={\tan }^{-1}\frac{8}{31}$
Taking tangent both sides :-
$\frac{(x+1)+(x-1)}{1-\left(x^2-1\right)}=\frac{8}{31}$
$\Rightarrow \frac{2x}{2-x^2}=\frac{8}{31}$
$\Rightarrow 4x^2+31x-8=0$
$\Rightarrow x=-8,\frac{1}{4}$
But, if $x=\frac{1}{4}$
${\tan }^{-1}(x+1)\in \left(0,\frac{\pi }{2}\right)$
$\&{\cot }^{-1}\left(\frac{1}{x-1}\right)\in \left(\frac{\pi }{2},\pi \right)$
$\Rightarrow LHS>\frac{\pi }{2}\&RHS<\frac{\pi }{2}$
(Not possible)
Hence, $x=-8$