3.120. $\operatorname{tg}^{2}\left(\alpha-\frac{3}{2} \pi\right)-\operatorname{ctg}^{2}\left(\alpha+\frac{3}{2} \pi\right)$.
## Solution.

$$
\begin{aligned}
& \operatorname{tg}^{2}\left(\alpha-\frac{3}{2} \pi\right)-\operatorname{ctg}^{2}\left(\alpha+\frac{3}{2} \pi\right)=\left(-\operatorname{tg}\left(\frac{3}{2} \pi-\alpha\right)\right)^{2}-\left(\operatorname{ctg}\left(\frac{3}{2} \pi+\alpha\right)\right)^{2}= \\
& =\operatorname{ctg}^{2} \alpha-\operatorname{tg}^{2} \alpha=\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}-\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}=\frac{\cos ^{4} \alpha-\sin ^{4} \alpha}{\sin ^{2} \alpha \cos ^{2} \alpha}= \\
& =\frac{4\left(\cos ^{2} \alpha-\sin ^{2} \alpha\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)\right.}{4 \sin ^{2} \alpha \cos ^{2} \alpha}=\frac{4 \cos 2 \alpha}{\sin ^{2} 2 \alpha}
\end{aligned}
$$

Answer: $4 \cos 2 \alpha \sin ^{-2} 2 \alpha$.