Question

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

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

Answer 1

The correct option is A $\frac{1}{2}$

$ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)$

$=ta{n}^{-1}\frac{(x+1+x-1)}{1-(x+1)(x-1)}$

$=ta{n}^{-1}\left(\frac{2x}{1-(x^2-1)}\right)$

$=ta{n}^{-1}\left(\frac{2x}{2-x^2}\right)=ta{n}^{-1}\left(\frac{4}{7}\right)$

$\therefore \frac{2x}{2-x^2}=\frac{4}{7}$

$\therefore 14x=8-4x^2$

$\therefore 4x^2+14x-8=0$

$\therefore 2x^2+7x-4=0$

$\therefore 2x^2+8x-x-4=0$

$\therefore 2x(x+4)-1(x+4)=0$

$\therefore x=-4$ OR $x=\frac{1}{2}$