Question

Solve for $x$:
${\tan }^{-1}(x+1)+{\tan }^{-1}(x-1)={\tan }^{-1}\frac{8}{31}$

Answer 1

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

We know that ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

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

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

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

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

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

$\Rightarrow \frac{x}{2-x^2}=\frac{4}{31}$

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

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

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

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

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

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

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

$x=-8$ is not a possible solution

$\therefore x=\frac{1}{4}$