3.322. $\frac{2 \cos ^{2}\left(\frac{9}{4} \pi-\alpha\right)}{1+\cos \left(\frac{\pi}{2}+2 \alpha\right)}-\frac{\sin \left(\alpha+\frac{7}{4} \pi\right)}{\sin \left(\alpha+\frac{\pi}{4}\right)} \cdot \cot\left(\frac{3}{4} \pi-\alpha\right)$.
Solution.

$$
\begin{aligned}
& \frac{2 \cos ^{2}\left(\frac{9}{4} \pi-\alpha\right)}{1+\cos \left(\frac{\pi}{2}+2 \alpha\right)}-\frac{\sin \left(\alpha+\frac{7}{4} \pi\right)}{\sin \left(\alpha+\frac{\pi}{4}\right)} \cdot \operatorname{ctg}\left(\frac{3}{4} \pi-\alpha\right)= \\
& =\frac{2 \cos ^{2}\left(\frac{9}{4} \pi-\alpha\right)}{1+\cos \left(\frac{\pi}{2}+2 \alpha\right)}-\frac{\sin \left(2 \pi-\left(\frac{\pi}{4}-\alpha\right)\right)}{\sin \left(\frac{\pi}{4}+\alpha\right)} \cdot \frac{1}{\operatorname{tg}\left(\frac{3}{4} \pi-\alpha\right)}= \\
& =\left[\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2} ; \operatorname{tg} \frac{x}{2}=\frac{\sin x}{1+\cos x}\right]= \\
& =\frac{1+\cos \left(\frac{9 \pi}{2}-2 \alpha\right)}{1-\sin 2 \alpha}-\frac{-\sin \left(\frac{\pi}{4}-\alpha\right)}{\sin \left(\frac{\pi}{4}+\alpha\right)} \cdot \frac{1}{\frac{\sin \left(\frac{3 \pi}{2}-2 \alpha\right)}{1+\cos \left(\frac{3 \pi}{2}-2 \alpha\right)}}=
\end{aligned}
$$

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

$$
=[\sin (x \pm y)=\sin x \cos y \pm \cos x \sin y]=
$$

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

$$
\begin{aligned}
& =\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}-\frac{(\cos \alpha-\sin \alpha)(1-\sin 2 \alpha)}{(\cos \alpha+\sin \alpha)(\cos \alpha-\sin \alpha)(\cos \alpha+\sin \alpha)}= \\
& =\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}-\frac{1-\sin 2 \alpha}{(\cos \alpha+\sin \alpha)^{2}}=\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}- \\
& -\frac{1-\sin 2 \alpha}{\cos ^{2} \alpha+2 \sin \alpha \cos \alpha+\sin ^{2} \alpha}= \\
& =\frac{1+\sin 2 \alpha}{1-\sin 2 \alpha}-\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}=\frac{(1+\sin 2 \alpha)^{2}-(1-\sin 2 \alpha)^{2}}{(1-\sin 2 \alpha)(1+\sin 2 \alpha)}= \\
& =\frac{1+2 \sin 2 \alpha+\sin ^{2} 2 \alpha-1+2 \sin 2 \alpha-\sin ^{2} 2 \alpha}{1-\sin ^{2} 2 \alpha}=\frac{4 \sin 2 \alpha}{\cos ^{2} 2 \alpha}
\end{aligned}
$$

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