$4.47 \frac{\sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)\left(\tan^{2} \alpha-1\right) \cot\left(\alpha-\frac{5 \pi}{4}\right)}{\sin ^{2}\left(\frac{5 \pi}{4}+\alpha\right)}=2$.
4.47 Note that

$$
\begin{aligned}
& \sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)=\cos ^{2} \alpha \\
& \operatorname{tg}^{2} \alpha-1=\frac{\sin ^{2} \alpha-\cos ^{2} \alpha}{\cos ^{2} \alpha}=\frac{(\sin \alpha-\cos \alpha)(\sin \alpha+\cos \alpha)}{\cos ^{2} \alpha} \\
& \sin \alpha-\cos \alpha=\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right) \\
& \sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\frac{\pi}{4}+\alpha\right) \\
& \operatorname{ctg}\left(\alpha-\frac{5 \pi}{4}\right)=\operatorname{ctg}\left(\alpha-\frac{\pi}{4}\right)
\end{aligned}
$$

Then the left side of the identity can be written as:

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

$$
\text { since } \cos \left(\alpha-\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}+\alpha\right) \text {. }
$$