3.099. $\frac{4 \sin ^{2}(\alpha-5 \pi)-\sin ^{2}(2 \alpha+\pi)}{\cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-4+4 \sin ^{2} \alpha}$.
## Solution.

$$
\begin{aligned}
& \frac{4 \sin ^{2}(\alpha-5 \pi)-\sin ^{2}(2 \alpha+\pi)}{\cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-4+4 \sin ^{2} \alpha}=\frac{4(-\sin (5 \pi-\alpha))^{2}-(\sin (\pi+2 \alpha))^{2}}{\left(\cos \left(\frac{3}{2} \pi-2 \alpha\right)\right)^{2}-4+4 \sin ^{2} \alpha}= \\
& =\frac{4 \sin ^{2} \alpha-\sin ^{2} 2 \alpha}{\sin ^{2} 2 \alpha-4+4 \sin ^{2} \alpha}=\frac{4 \sin ^{2} \alpha-4 \sin ^{2} \alpha \cos ^{2} \alpha}{4 \sin ^{2} \alpha \cos ^{2} \alpha-4+4\left(1-\cos ^{2} \alpha\right)}= \\
& =\frac{4 \sin ^{2} \alpha\left(1-\cos ^{2} \alpha\right)}{4\left(1-\cos ^{2} \alpha\right) \cos ^{2} \alpha-4+4-4 \cos ^{2} \alpha}=\frac{4 \sin ^{2} \alpha \sin ^{2} \alpha}{4 \cos ^{4} \alpha-4 \cos ^{2} \alpha}= \\
& =\frac{4 \sin ^{4} \alpha}{4 \cos ^{4} \alpha}=-\tan^{4} \alpha
\end{aligned}
$$

Answer: $-\tan^{4} \alpha$.