If tan frac{A}{2} = frac{3}{2}, then frac{1 + | vedclass

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(D) Given that $	an \frac{A}{2} = \frac{3}{2}$.We know the trigonometric identities: $1 + \cos A = 2 \cos^2 \frac{A}{2}$ and $1 - \cos A = 2 \sin^2 \

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$4/9$

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(D) Given that $	an \frac{A}{2} = \frac{3}{2}$.We know the trigonometric identities: $1 + \cos A = 2 \cos^2 \frac{A}{2}$ and $1 - \cos A = 2 \sin^2 \frac{A}{2}$.Substituting these into the expression:$\frac{1 + \cos A}{1 - \cos A} = \frac{2 \cos^2 \frac{A}{2}}{2 \sin^2 \frac{A}{2}}$$= \cot^2 \frac{A}{2}$Since $	an \frac{A}{2} = \frac{3}{2}$,then $\cot \frac{A}{2} = \frac{1}{	an \frac{A}{2}} = \frac{2}{3}$.Therefore,$\cot^2 \frac{A}{2} = \left( \frac{2}{3} ight)^2 = \frac{4}{9}$.

If $\tan \frac{A}{2} = \frac{3}{2},$ then $\frac{1 + \cos A}{1 - \cos A} = $

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