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