If \(f(x)=\frac{1+x}{1-x}\), then \(\frac{f(x).f(x^2)}{1+[f(x)]^2}\) is equal to

1
\(\frac{1}{4}\)
2
\(\frac{1}{6}\)
3
\(\frac{1}{8}\)
4
\(\frac{1}{2}\)

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