Let f : (-1, 1) → B, be a function defined by $f(x)=\tan^{-1}\frac{2x}{1-x^2}$, then f is both one-one and onto when B is the interval.

$(-\frac{π}{2},\frac{π}{2})$

We know that $y=\tan^{-1}(\frac{2x}{1-x^2})=2\tan^{-1}(x)$

y is an increasing function as $y'=\frac{2}{1-x^2}>0$

so $y_{min}=y(-1)=2\tan^{-1}(-1)=-\frac{π}{2}$

as $x → -1,\,\, y → -\frac{π}{2}$

as $x → 1,\,\, y → \frac{π}{2}$

so $B= (-\frac{π}{2},\frac{π}{2})$