3.076. $\frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}=\frac{\left(1-1+2 \sin ^{2} \alpha\right) \cdot \frac{\sqrt{2}}{2}(\cos 2 \alpha-\sin 2 \alpha)}{2 \sin ^{2} 2 \alpha-2 \sin 2 \alpha \cos 2 \alpha}= \\
& =\frac{\sqrt{2} \sin ^{2} \alpha(\cos 2 \alpha-\sin 2 \alpha)}{-2 \sin 2 \alpha(\cos 2 \alpha-\sin 2 \alpha)}=\frac{\sqrt{2} \sin ^{2} \alpha}{-2 \sin 2 \alpha}=-\frac{\sqrt{2} \sin ^{2} \alpha}{4 \sin \alpha \cos \alpha}=-\frac{\sqrt{2} \sin \alpha}{4 \cos \alpha}= \\
& =-\frac{\sqrt{2}}{4} \tan \alpha
\end{aligned}
$$

Answer: $-\frac{\sqrt{2}}{4} \tan \alpha$.