B

- Hướng dẫn giải

Vì $\mathrm{\pi }<\mathrm{a}<\frac{3\mathrm{\pi }}{2}$ nên  sina < 0; cosa < 0. Ta có

$$\begin{gathered} \left\{\begin{array}{l}\sin \left(\alpha \right)-2\cos \left(\alpha \right)=1\\ {\sin }^2\left(\alpha \right)+{\cos }^2\left(\alpha \right)=1\end{array}\right. \\ \Rightarrow {\left(1+2\cos \left(\alpha \right)\right)}^2+{\cos }^2\left(\alpha \right)=1 \\ \Rightarrow 5{\cos }^2\left(\alpha \right)+4\cos \left(\alpha \right)=0 \\ \Rightarrow \cos \left(\alpha \right)=-\frac{4}{5} \end{gathered}$$ 

Suy ra $\alpha =-\sqrt{1-{\cos }^2\left(\alpha \right)}=-\frac{3}{5}$; $\tan \left(\alpha \right)=\frac{3}{4}$; $cot\left(\alpha \right)=\frac{4}{3}$. Vậy A = 2tana - cota = $2.\frac{3}{4}-\frac{4}{3}=\frac{1}{6}$

Đáp án B