Bài tập 6.18 trang 185 SBT Toán 10

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 HỌC247