Cho pi < alpha <3pi /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.