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 C $\frac{1}{4}$

We have, ${\tan }^{-1}\left[\frac{x+1+x-1}{1-(x+1)(x-1)}\right]={\tan }^{-1}\frac{8}{31}$
$\Rightarrow \frac{2x}{1-(x^2-1)}=\frac{8}{31}\Rightarrow 62x=16-8x^2$
$\Rightarrow 8x^2+62x-16=0\Rightarrow 4x^2+31x-8=0$
$\Rightarrow 4x^2+32x-x-8=0\Rightarrow 4x(x-8)-1(x+8)=0$
$\Rightarrow (x+8)(4x-1)=0\Rightarrow x=-8or\frac{1}{4}$
$\therefore x=\frac{1}{4}$
Neglecting $x=-8$ as $x^2-1<1$