lf $2 t a n h^{-} 1 =$ logy,then the value of $y$ in terms of $x$ is

2 x

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\frac{2 x}{1 - x^{2}}

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X^{2}

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(\frac{1 + x}{1 - x})

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Solution

The correct option is D $(\frac{1 + x}{1 - x})$
$tan h^{- 1} x = \frac{1}{2} ln (\frac{1 + x}{1 - x})$ [formula]
$2 tan h^{- 1} (x) = ln (\frac{1 + x}{1 - x}) = ln y$ [Given]
$y = \frac{1 + x}{1 - y}$

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