3.101. $\operatorname{ctg}(4 \alpha-\pi)\left(\cos ^{4}\left(\frac{5}{4} \pi-2 \alpha\right)-\sin ^{4}\left(\frac{9}{4} \pi-2 \alpha\right)\right)$.
Solution.

$$
\begin{aligned}
& \operatorname{ctg}(4 \alpha-\pi)\left(\cos ^{4}\left(\frac{5}{4} \pi-2 \alpha\right)-\sin ^{4}\left(\frac{9}{4} \pi-2 \alpha\right)\right)= \\
& =-\operatorname{ctg}(\pi-4 \alpha)\left(\left(\cos \left(\frac{4 \pi+\pi}{4}-2 \alpha\right)\right)^{4}-\left(\sin \left(\frac{8 \pi+\pi}{4}-2 \alpha\right)\right)^{4}\right)= \\
& =-\operatorname{ctg}(\pi-4 \alpha)\left(\left(\cos \left(\pi+\left(\frac{\pi}{4}-2 \alpha\right)\right)\right)^{4}-\left(\sin \left(2 \pi+\left(\frac{\pi}{4}-2 \alpha\right)\right)\right)^{4}\right)= \\
& =\operatorname{ctg} 4 \alpha\left(\cos ^{4}\left(\frac{\pi}{4}-2 \alpha\right)-\sin ^{4}\left(\frac{\pi}{4}-2 \alpha\right)\right)= \\
& =\operatorname{ctg} 4 \alpha\left(\left(\cos ^{2}\left(\frac{\pi}{4}-2 \alpha\right)\right)^{2}-\left(\sin ^{2}\left(\frac{\pi}{4}-2 \alpha\right)\right)^{2}\right)=
\end{aligned}
$$

$$
=\operatorname{ctg} 4 \dot{\alpha} \cos \left(\frac{\pi}{2}-4 \alpha\right)=\frac{\cos 4 \alpha}{\sin 4 \alpha} \cdot \sin 4 \alpha=\cos 4 \alpha
$$

Answer: $\cos 4 \alpha$.

$$
\text { 3.102. } \frac{\cos ^{2}\left(\frac{5}{4} \pi-2 \alpha\right)-\sin ^{2}\left(\frac{5}{4} \pi-2 \alpha\right)}{\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)\left(\cos \left(2 \pi-\frac{\alpha}{2}\right)+\cos \left(\frac{\pi}{2}+\frac{\alpha}{2}\right)\right) \sin \alpha} \text {. }
$$

## Solution.

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

Answer: $4 \cos 2 \alpha$.