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In triangle $\mathrm{A B C}$, altitudes $\mathrm{A D}$ and $\mathrm{B E}$ are drawn to the corresponding bases. If $\angle \mathrm{B A C}=45^{\circ}$ and $\angle \mathrm{A B C}=\theta$, then $\frac{A D}{B E}$ equals

- $\sqrt{2} \cos \theta$
- $1$
- $\sqrt{2} \sin \theta$
- $\frac{(\sin +\cos )}{\sqrt{2}}$