Description for Correct answer:


Since, \( \Large f: \left(-1,\ 1\right) \rightarrow B \)

Let \( \Large x\ \epsilon\ \left(-1,\ 1\right) \)

=> \( \Large \tan^{-1}\ x\ \epsilon\ \left(-\frac{ \pi }{4},\ \frac{ \pi }{4}\right) => 2 \tan^{-1} x\ \epsilon\ \left(-\frac{ \pi }{2},\ \frac{ \pi }{2}\right) \)

and \( \Large f \left(x\right) = \tan^{-1} \frac{2x}{1-x^{2}}=2 \tan^{-1}x,\ \left(x^{2} < 1\right) \)

So \( \Large f \left(x\right)\ \epsilon\ \left(-\frac{ \pi }{2},\ \frac{ \pi }{2}\right) \)

Therefore, Function is one-one onto.