Question

The sum of possible values of $x$ for ${\tan }^{-1}(x+1)+co{t}^{-1}\frac{1}{(x-1)}={\tan }^{-1}\frac{8}{31}$ is

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

Answer 1

The correct option is C

$\frac{–31}{4}$

Explanation for the correct option:

Find the required sum.

Given:${\tan }^{-1}(x+1)+co{t}^{-1}\frac{1}{(x-1)}={\tan }^{-1}\frac{8}{31}$

$\begin{array}{rcl}& \Rightarrow & {\tan }^{-1}(x+1)+{\tan }^{-1}(x-1)={\tan }^{-1}\frac{8}{31}\left[\because co{t}^{-1}\frac{1}{\left(x-1\right)}={\tan }^{-1}(x-1)\right]\\ & \Rightarrow & {\tan }^{-1}\left(\frac{x+1+x-1}{1-(x+1)(x-1)}\right)={\tan }^{-1}\frac{8}{31}\left[\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)\right]\\ & \Rightarrow & \frac{2x}{1-\left(x^2-1^2\right)}=\frac{8}{31}\\ & \Rightarrow & \frac{2x}{2-x^2}=\frac{8}{31}\\ & \Rightarrow & 62x=16-8x^2\\ & \Rightarrow & 8x^2+62x-16=0\\ & \Rightarrow & 4x^2+31x-8=0\\ & \Rightarrow & 4x^2+32x-x-8=0\\ & \Rightarrow & 4x\left(x+8\right)-1(x+8)=0\\ & \Rightarrow & \left(4x-1\right)\left(x+8\right)=0\\ & \Rightarrow & x=\frac{1}{4}orx=-8\end{array}$

So, the sum of possible values of $x$ is $\frac{1}{4}+\left(–8\right)=\frac{-31}{4}$

Hence, option (C) is the correct answer.