Question

If $ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)=ta{n}^{-1}\left(\frac{8}{31}\right)$, 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, $ta{n}^{-1}\left[\frac{x+1+x-1}{1(x+1)(x-1)}\right]=ta{n}^{-1}\frac{8}{31}$
$\Rightarrow \frac{2x}{1-(x^2-1)}=\frac{8}{31}\Rightarrow 62x=16-8x^2$
$\Rightarrow 8x^2+62x-16=0\Rightarrow 4x^2+31x-8=0$
$\Rightarrow 4x^2+32x-x-8=0\Rightarrow 4x(x-8)-1(x+8)=0$
$\Rightarrow (x+8)(4x-1)=0\Rightarrow x=-8.\frac{1}{4}$
$\therefore x=\frac{1}{4}.Neglectingx=-8.as.x^2-1<1$