If $\frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}} = \frac{x}{y}$, then the value of $\tan A$ is:

Question

$\frac{x^2 + y^2}{x^2 - y^2}$

$\frac{2xy}{x^2 + y^2}$

$\frac{2xy}{x^2 - y^2}$

$\frac{2xy}{y^2 - x^2}$

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