Cho \({\rm{cos}}\alpha  =  - \frac{2}{5}\,\,\,\lef...

B

- Hướng dẫn giải

Ta có: \(\sin \alpha  = \sqrt {1 - c{\rm{o}}{{\rm{s}}^2}\alpha }  =  \pm \frac{{\sqrt {21} }}{5}\)

Vì \({\pi  < \alpha  < \frac{{2\pi }}{3}}\) nên \(\sin \alpha  =  - \frac{{\sqrt {21} }}{5}\)

Suy ra \(\tan \alpha  = \frac{{\sin \alpha }}{{{\rm{cos}}\alpha }} = \frac{{ - \frac{{\sqrt {21} }}{5}}}{{\frac{{ - 2}}{5}}} =  - \frac{{\sqrt {21} }}{2}\)