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$\frac{\sin 2 A}{1+\cos 2 A} \cdot \frac{\cos A}{1+\cos A}=$
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Verified Answer
The correct answer is:
$\tan \frac{A}{2}$
$\begin{aligned}
& \left(\frac{\sin 2 A}{1+\cos 2 A}\right)\left(\frac{\cos A}{1+\cos A}\right) \\
& =\frac{2 \sin A \cos A}{2 \cos ^2 A} \frac{\cos A}{1+\cos A}=\frac{\sin A}{1+\cos A}=\tan \frac{A}{2}
\end{aligned}$
& \left(\frac{\sin 2 A}{1+\cos 2 A}\right)\left(\frac{\cos A}{1+\cos A}\right) \\
& =\frac{2 \sin A \cos A}{2 \cos ^2 A} \frac{\cos A}{1+\cos A}=\frac{\sin A}{1+\cos A}=\tan \frac{A}{2}
\end{aligned}$
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