3.320. $\frac{1-2 \sin ^{2} \alpha}{2 \operatorname{tg}\left(\frac{5}{4} \pi+\alpha\right) \cos ^{2}\left(\frac{\pi}{4}+\alpha\right)}-\operatorname{tg} \alpha+\sin \left(\frac{\pi}{2}+\alpha\right)-\cos \left(\alpha-\frac{\pi}{2}\right)$.
Solution.

$$
=\frac{1-2 \sin ^{2} \alpha}{\operatorname{tg}\left(\frac{5}{4} \pi+\alpha\right)\left(2 \cos ^{2}\left(\frac{\pi}{4}+\alpha\right)\right)}-\operatorname{tg} \alpha+\sin \left(\frac{\pi}{2}+\alpha\right)-\cos \left(\frac{\pi}{2}-\alpha\right)=
$$

$$
\begin{aligned}
& =\left[1-2 \sin ^{2} x=\cos 2 x, \operatorname{tg} \frac{x}{2}=\frac{\sin x}{1+\cos x}, x \neq \pi+2 \pi n, n \in Z\right. \\
& \left.\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}\right]= \\
& =\frac{\cos 2 \alpha}{\frac{\sin \left(\frac{5}{2} \pi+2 \alpha\right)}{1+\cos \left(\frac{5}{2} \pi+2 \alpha\right)} \cdot\left(1+\cos \left(\frac{\pi}{2}+2 \alpha\right)\right)}-\operatorname{tg} \alpha+\cos \alpha-\sin \alpha= \\
& =\frac{\cos 2 \alpha}{\frac{\cos 2 \alpha}{1-\sin 2 \alpha} \cdot(1-\sin 2 \alpha)}-\frac{\sin \alpha}{\cos \alpha}+\cos \alpha-\sin \alpha= \\
& =\left(1-\frac{\sin \alpha}{\cos \alpha}\right)+(\cos \alpha-\sin \alpha)=\frac{\cos \alpha-\sin \alpha}{\cos \alpha}+(\cos \alpha-\sin \alpha)= \\
& =(\cos \alpha-\sin \alpha)\left(\frac{1}{\cos \alpha}+1\right)=\frac{(\cos \alpha-\sin \alpha)(1+\cos \alpha)}{\cos \alpha}= \\
& =\frac{\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \cos \alpha-\frac{\sqrt{2}}{2} \sin \alpha\right)\left(1+\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)}{\cos \alpha}= \\
& =\frac{\sqrt{2}\left(\cos \frac{\pi}{4} \cos \alpha-\sin \frac{\pi}{4} \sin \alpha\right) \cdot 2 \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}= \\
& =\frac{2 \sqrt{2}\left(\cos \frac{\pi}{4} \cos \alpha-\sin \frac{\pi}{4} \sin \alpha\right) \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}= \\
& =[\cos x \cos y-\sin x \sin y=\cos (x+y)]=\frac{2 \sqrt{2} \cos \left(\frac{\pi}{4}+\alpha\right) \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}
\end{aligned}
$$

Answer: $\frac{2 \sqrt{2} \cos \left(\frac{\pi}{4}+\alpha\right) \cos ^{2} \frac{\alpha}{2}}{\cos \alpha}$.