Cho pi < α < 3 pi/ 2 . Khẳng định nào sau đây đúng?

\(\tan \left( {\pi - \alpha } \right) = - \tan \alpha < 0\) vì \(\pi < \alpha < \frac{{3\pi }}{2}\)\( \Rightarrow \tan \alpha > 0\).

Có \(\pi < \alpha < \frac{{3\pi }}{2}\)\( \Rightarrow \frac{{5\pi }}{2} < \frac{{3\pi }}{2} + \alpha < 3\pi \)\( \Rightarrow \tan \left( {\frac{{3\pi }}{2} + \alpha } \right) < 0\).

Ta có \(\pi < \alpha < \frac{{3\pi }}{2}\)\( \Rightarrow - \frac{{3\pi }}{2} < - \alpha < - \pi \)\( \Rightarrow 0 < \frac{{3\pi }}{2} - \alpha < \frac{{3\pi }}{2} - \pi \)\( \Rightarrow 0 < \frac{{3\pi }}{2} - \alpha < \frac{\pi }{2}\).

Do đó \(\sin \left( {\frac{{3\pi }}{2} - \alpha } \right) > 0;\cos \left( {\frac{{3\pi }}{2} - \alpha } \right) > 0 \Rightarrow \tan \left( {\frac{{3\pi }}{2} - \alpha } \right) > 0\). Chọn B.