Cho \(\tan \alpha - 3\cot \alpha = 6\) và \(\pi < \alpha < \frac{{3\pi }}{2}\). Tính
a) \(\sin \alpha + \cos \alpha \)
b) \(\frac{{2\sin \alpha - \tan \alpha }}{{\cos \alpha + \cot \alpha }}\)
Vì \(\pi < \alpha < \frac{{3\pi }}{2}\) nên \(\cos \alpha < 0,\sin \alpha < 0\) và \(\tan \alpha > 0\)
Ta có \[\tan \alpha - 3\cot \alpha = 6 \Leftrightarrow \tan \alpha - \frac{3}{{\tan \alpha }} - 6 = 0 \Leftrightarrow {\tan ^2}\alpha - 6\tan \alpha - 3 = 0\)
Vì \(\tan \alpha > 0\) nên \(\tan \alpha = 3 + 2\sqrt 3 \)
a) \({\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{1}{{22 + 12\sqrt 3 }}\)
Suy ra \(\cos \alpha = - \frac{1}{{\sqrt {22 + 12\sqrt 3 } }},\sin \alpha = - \frac{{3 + 2\sqrt 3 }}{{\sqrt {22 + 12\sqrt 3 } }}\)
Vậy \(\sin \alpha + \cos \alpha = - \frac{{4 + 2\sqrt 3 }}{{\sqrt {22 + 12\sqrt 3 } }}\)
b) \(\frac{{2\sin \alpha - \tan \alpha }}{{\cos \alpha + \cot \alpha }} = \frac{{\sin \alpha \left( {2 - \frac{1}{{\cos \alpha }}} \right)}}{{\cos \alpha \left( {1 + \frac{1}{{\sin \alpha }}} \right)}} = \tan \alpha .\frac{{2\cos \alpha - 1}}{{\cos \alpha }}.\frac{{\sin \alpha }}{{\sin \alpha + 1}} = {\tan ^2}\alpha .\frac{{2\cos \alpha - 1}}{{\sin \alpha + 1}}\)
\( = \left( {21 + 12\sqrt 3 } \right).\frac{{2 + \sqrt {22 + 12\sqrt 3 } }}{{3 + 2\sqrt 2 - \sqrt {22 + 12\sqrt 3 } }}\)
-- Mod Toán 10