Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
cos\alpha < 0
\end{array} \right.\\
\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \dfrac{{2\sqrt 6 }}{7}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = - \dfrac{{2\sqrt 6 }}{5}\\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{ - 5}}{{2\sqrt 6 }}\\
7,\\
\dfrac{\pi }{2} < x < \pi \Rightarrow \left\{ \begin{array}{l}
\sin x > 0\\
\cos x < 0
\end{array} \right.\\
\sin x > 0 \Rightarrow \sin x = \sqrt {1 - {{\cos }^2}x} = \dfrac{{2\sqrt 2 }}{3}\\
\tan x = \dfrac{{\sin x}}{{\cos x}} = - 2\sqrt 2 \\
\cot x = \dfrac{{\cos x}}{{\sin x}} = - \dfrac{{\sqrt 2 }}{4}\\
4,\\
0 < \alpha < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha > 0
\end{array} \right.\\
\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } = \dfrac{{\sqrt {11} }}{4}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{\sqrt 5 }}{{\sqrt {11} }}\\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{\sqrt {11} }}{{\sqrt 5 }}
\end{array}\)
