4.40 Find $\operatorname{ctg} 2 \alpha$, if it is known that $\sin \left(\alpha-90^{\circ}\right)=-\frac{2}{3}$ and $270^{\circ}<\alpha<360^{\circ}$.
4.40 We have $\sin \left(\alpha-90^{\circ}\right)=-\sin \left(90^{\circ}-\alpha\right)=-\frac{2}{3}$, i.e., $\cos \alpha=\frac{2}{3}$.

Therefore, $\sin \alpha=-\sqrt{1-\frac{4}{9}}=-\frac{\sqrt{5}}{3}$ (since $\alpha$ is an angle in the IV quadrant). Next, we find

$$
\operatorname{ctg} 2 \alpha=\frac{\cos 2 \alpha}{\sin 2 \alpha}=\frac{2 \cos ^{2} \alpha-1}{2 \sin \alpha \cos \alpha}=\frac{2 \cdot \frac{4}{9}-1}{2 \cdot \frac{2}{3}\left(-\frac{\sqrt{5}}{3}\right)}=\frac{\sqrt{5}}{20}
$$