Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \dfrac{{\sqrt 5 }}{3}\\
\tan a = \dfrac{{\sin a}}{{\cos a}} = - \dfrac{2}{{\sqrt 5 }}\\
\cot a = \dfrac{{\cos a}}{{\sin a}} = - \dfrac{{\sqrt 5 }}{2}\\
b,\\
P = \dfrac{{\sin x + \sin 3x + \sin 5x}}{{\cos x + \cos 3x + \cos 5x}}\\
= \dfrac{{\left( {\sin x + \sin 5x} \right) + \sin 3x}}{{\left( {\cos x + \cos 5x} \right) + \cos 3x}}\\
= \dfrac{{2\sin 3x.\cos 2x + \sin 3x}}{{2.\cos 3x.\cos 2x + \cos 3x}}\\
= \dfrac{{\sin 3x\left( {2\cos 2x + 1} \right)}}{{\cos 3x\left( {2\cos 2x + 1} \right)}}\\
= \dfrac{{\sin 3x}}{{\cos 3x}} = \tan 3x\\
2,\\
x \ge 5 \Rightarrow x > 4 \Rightarrow 4 - x < 0,\,\,\,\,\forall x \ge 5\\
3,\\
\dfrac{{3x - 6}}{{5 + x}} \le 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 5\\
\left( {3x - 6} \right)\left( {x + 5} \right) \le 0
\end{array} \right. \Leftrightarrow - 5 < x \le 2\\
x \in Z \Rightarrow x \in \left\{ { - 4; - 3; - 2; - 1;0;1;2} \right\}
\end{array}\)
