Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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