$4.60 \frac{4 \sin \left(4 \alpha-\frac{\pi}{2}\right)}{\operatorname{ctg}^{2}\left(2 \alpha-\frac{3 \pi}{2}\right)-\operatorname{tg}^{2}\left(2 \alpha+\frac{5 \pi}{2}\right)}-1$.
4.60 We have

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

$=\frac{-\cos 4 \alpha \sin ^{2} 4 \alpha}{-\cos 4 \alpha}-1=\sin ^{2} 4 \alpha-1=-\cos ^{2} 4 \alpha$.