Let $f:(-1,1)\to B$ 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:

The function is $f(x)={\tan }^{-1}\frac{2x}{1-x^2}$.This is a standard identity: ${\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x$ for $x\in (-1,1)$.The range of $y={\tan }^{-1}x$ for $x\in (-1,1)$ is $(-\pi /4,\pi /4)$.Therefore, the range of $f(x)=2{\tan }^{-1}x$ is:$2\times (-\pi /4,\pi /4)=(-\pi /2,\pi /2)$.For the function to be onto, the co-domain $B$ must be equal to its range.Thus, $B=(-\pi /2,\pi /2)$.