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

$$
\begin{aligned}
& \sin \left(\alpha-90^{\circ}\right)=-\frac{2}{3}, \quad-\sin \left(90^{\circ}-\alpha\right)=-\frac{2}{3}, \quad \sin \left(90^{\circ}-\alpha\right)=\frac{2}{3} \\
& \cos \alpha=\frac{2}{3}, \quad \cos ^{2} \alpha=\frac{4}{9} \\
& 1-\sin ^{2} \alpha=\frac{4}{9}, \quad \sin ^{2} \alpha=\frac{5}{9}
\end{aligned}
$$

Considering that $x \in\left(270^{\circ} ; 360^{\circ}\right)$, we have $\sin \alpha=-\frac{\sqrt{5}}{3}$. Taking into account the found values, we get

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

Answer: $\frac{\sqrt{5}}{20}$.