Question

$\text{If}{\tan }^{-1}(x+1)+{\tan }^{-1}(x-1)={\tan }^{-1}\left(\frac{8}{31}\right),\text{then}x\text{is}\text{equal}$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{4}$
• D. 1

Answer 1

The correct option is C $\frac{1}{4}$

${\tan }^{-1}(x+1)+{\tan }^{-1}\frac{8}{31}$

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

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

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

$\Rightarrow 62x=16-8x^2$

$\Rightarrow 8x^2+62x-16=0$

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

$\Rightarrow 4x^2+32x-8=0$

$\Rightarrow 4x(x+8)-1(x+8)=0$

$\Rightarrow (4x-1)(x+8)=0$

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