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

Answer 1

The correct option is B $-\frac{32}{4}$

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

Checking the values,
At $x=\frac{1}{4}$
$L.H.S.>\frac{\pi }{2}\text{and}R.H.S.<\frac{\pi }{2}$
So, the only possible solution is $x=-8$
Hence, sum of solution $=-\frac{32}{4}$