Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\pi < \alpha < \frac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha < 0
\end{array} \right.\\
\tan \alpha - 3\cot \alpha = 6\\
\Leftrightarrow tan\alpha - \frac{3}{{\tan \alpha }} = 6\\
\Leftrightarrow {\tan ^2}\alpha - 6\tan \alpha - 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan \alpha = 3 + 2\sqrt 3 \\
\tan \alpha = 3 - 2\sqrt 3
\end{array} \right.\\
\sin \alpha > 0,\,\,\,\cos \alpha > 0 \Rightarrow \tan \alpha = 3 + 2\sqrt 3 \\
\Rightarrow \frac{{\sin \alpha }}{{\cos \alpha }} = 3 + 2\sqrt 3 \Leftrightarrow \sin \alpha = \left( {3 + 2\sqrt 3 } \right)\cos \alpha \\
a,\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( {3 + 2\sqrt 3 } \right)^2}{\cos ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \frac{{11 - 6\sqrt 3 }}{{26}}\\
\Rightarrow \cos \alpha = - \sqrt {\frac{{11 - 6\sqrt 3 }}{{26}}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\cos \alpha < 0} \right)\\
\Rightarrow \sin \alpha = ...\\
b,\\
\frac{{2\sin \alpha - \tan \alpha }}{{\cos \alpha + \cot \alpha }} = \frac{{2\sin \alpha - \frac{{\sin \alpha }}{{\cos \alpha }}}}{{\cos \alpha + \frac{{\cos \alpha }}{{\sin \alpha }}}} = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{2 - \frac{1}{{\cos \alpha }}}}{{1 + \frac{1}{{\sin \alpha }}}}\\
= \tan \alpha .\frac{{\frac{{2\cos \alpha - 1}}{{\cos \alpha }}}}{{\frac{{\sin \alpha + 1}}{{\sin \alpha }}}} = {\tan ^2}\alpha .\frac{{2\cos \alpha - 1}}{{\sin \alpha + 1}} = .....
\end{array}\)
