Giúp mk vs mai mk phải nộp rồi

Giải thích các bước giải:

 Dựa vào đường tròn lượng giắc ta có:

\(\begin{array}{l}
0 < x < \frac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x > 0
\end{array} \right.\\
\frac{\pi }{2} < x < \pi  \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x < 0
\end{array} \right.\\
\pi  < x < \frac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
\frac{{3\pi }}{2} < x < 2\pi  \Rightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x > 0
\end{array} \right.\\
a,\\
1,\\
\frac{\pi }{2} < \alpha  < \pi  \Rightarrow 0 < \alpha  - \frac{\pi }{2} < \frac{\pi }{2} \Rightarrow \sin \left( {\alpha  - \frac{\pi }{2}} \right) > 0\\
2,\\
\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \pi  < \alpha  + \frac{\pi }{2} < \frac{{3\pi }}{2} \Rightarrow \cos \left( {\alpha  + \frac{\pi }{2}} \right) < 0\\
3,\\
\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \frac{{3\pi }}{2} < \alpha  + \pi  < 2\pi  \Rightarrow \left\{ \begin{array}{l}
\sin \left( {\alpha  + \pi } \right) < 0\\
\cos \left( {\alpha  + \pi } \right) > 0
\end{array} \right. \Rightarrow \tan \left( {\alpha  + \pi } \right) < 0\\
4,\\
\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \frac{\pi }{2} < \frac{{3\pi }}{2} - \alpha  < \pi  \Rightarrow \left\{ \begin{array}{l}
\sin \left( {\frac{{3\pi }}{2} - \alpha } \right) > 0\\
\cos \left( {\frac{{3\pi }}{2} - \alpha } \right) < 0
\end{array} \right. \Rightarrow \cot \left( {\frac{{3\pi }}{2} - \alpha } \right) < 0\\
b,\\
1,\\
0 < \alpha  < \frac{\pi }{2} \Rightarrow  - \pi  < \alpha  - \pi  <  - \frac{\pi }{2} \Leftrightarrow \pi  < \alpha  - \pi  < \frac{{3\pi }}{2} \Rightarrow \sin \left( {\alpha  - \pi } \right) < 0\\
2,\\
0 < \alpha  < \frac{\pi }{2} \Rightarrow \frac{{3\pi }}{2} < \alpha  + \frac{{3\pi }}{2} < 2\pi  \Rightarrow \cos \left( {\alpha  + \frac{{3\pi }}{2}} \right) > 0\\
3,\\
0 < \alpha  < \frac{\pi }{2} \Rightarrow \frac{\pi }{2} < \alpha  + \frac{\pi }{2} < \pi  \Rightarrow \left\{ \begin{array}{l}
\sin \left( {\alpha  + \frac{\pi }{2}} \right) > 0\\
\cos \left( {\alpha  + \frac{\pi }{2}} \right) < 0
\end{array} \right. \Rightarrow \tan \left( {\alpha  + \frac{\pi }{2}} \right) < 0\\
4,\\
0 < \alpha  < \frac{\pi }{2} \Rightarrow \pi  < \alpha  + \pi  < \frac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin \left( {\alpha  + \pi } \right) < 0\\
\cos \left( {\alpha  + \pi } \right) < 0
\end{array} \right. \Rightarrow \cot \left( {\alpha  + \pi } \right) > 0
\end{array}\)