C

- Hướng dẫn giải

Ta có

$$\begin{gathered} \frac{1}{{\cos }^2\left(\alpha \right)}=1+{\tan }^2\left(\alpha \right) \\ =1+4=5 \end{gathered}$$

Vì $\mathrm{\pi }<\mathrm{\alpha }<\frac{3\mathrm{\pi }}{2}$ nên $\cos \left(\alpha \right)<0$

Suy ra $\cos \left(\alpha \right)=\frac{1}{\sqrt{5}}$

Khi đó

$M={\sin }^2\left(\alpha \right)+\sin \left(\alpha +\frac{\mathrm{\pi }}{2}\right)+\sin \left(\frac{5\mathrm{\pi }}{2}-2\alpha \right)$

$$\begin{gathered} ={\sin }^2\left(\alpha \right)+\cos \left(\alpha \right)+\cos \left(2\alpha \right) \\ ={\sin }^2\left(\alpha \right)+\cos \left(\alpha \right)+2{\cos }^2\left(\alpha \right)-1 \\ ={\cos }^2\left(\alpha \right)+\cos \left(\alpha \right) \\ =\frac{1}{5}-\frac{1}{\sqrt{5}}=\frac{1-\sqrt{5}}{5} \end{gathered}$$

Đáp án C