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Let $f:(-1,1) \rightarrow B$, be a function defined by $f(x)=\tan ^{-1} \frac{2 x}{1-x^2}$, then $f$ is both one-one and onto when $B$ is the interval
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$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$
Given $f(x)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ for $x \in(-1,1)$
clearly range of $f(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\therefore$ co-domain of function $=B=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
clearly range of $f(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
$\therefore$ co-domain of function $=B=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
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