3.174. Find $\operatorname{tg} 2 \alpha$, if it is known that $\cos \left(\alpha-90^{\circ}\right)=0.2$ and $90^{\circ}<\alpha<180^{\circ}$.
## Solution.

$$
\begin{aligned}
& \cos \left(\alpha-90^{\circ}\right)=\cos \left(90^{\circ}-\alpha\right)=\sin \alpha=0.2, \sin ^{2} \alpha=0.04 \\
& 1-\cos ^{2} \alpha=0.04, \cos ^{2} \alpha=0.96=\frac{24}{25}
\end{aligned}
$$

$$
\begin{aligned}
& \cos \alpha=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5} \text { when } \alpha \in\left(90^{\circ} ; 180^{\circ}\right) . \\
& \text { Then, } \operatorname{tg} 2 \alpha=\frac{\sin 2 \alpha}{\cos 2 \alpha}=\frac{2 \sin \alpha \cos \alpha}{\cos ^{2} \alpha-\sin ^{2} \alpha}=\frac{2 \cdot \frac{1}{5} \cdot\left(-\frac{2 \sqrt{6}}{5}\right)}{\frac{24}{25}-\frac{1}{25}}=-\frac{4 \sqrt{6}}{23} .
\end{aligned}
$$

Answer: $-\frac{4 \sqrt{6}}{23}$.