9. Simplify $\frac{\sin 4 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \tan \left(\frac{\pi}{4}-\alpha\right)}=(\quad)$.
(A) $\cos 2 \alpha$
(B) $\sin 2 \alpha$
(C) $\cos \alpha$
(D) $\sin \alpha$
9. B.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \tan \left[\frac{\pi}{2}-\left(\frac{\pi}{4}+\alpha\right)\right]} \\
=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{4 \sin ^{2}\left(\frac{\pi}{4}+\alpha\right) \cdot \cot \left(\frac{\pi}{4}+\alpha\right)} \\
=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{4 \sin \left(\frac{\pi}{4}+\alpha\right) \cdot \cos \left(\frac{\pi}{4}+\alpha\right)} \\
=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{2 \sin \left(\frac{\pi}{2}+2 \alpha\right)}=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{2 \cos 2 \alpha} \\
=\sin 2 \alpha .
\end{array}
$$