Question

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

• A. $\frac{1}{4}$
• B. $\frac{2}{4}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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