Cho góc α thỏa mãn pi < α < 3 pi/ 2 và sin α − 2 cos α = 1 . Tính P = 2 tan ( α + 5 pi) + cot ( 3pi − α ) .

Chọn C

Vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\) nên \(\sin \alpha  < 0,\cos \alpha  < 0\).

Ta có: \(\left\{ \begin{array}{l}\sin \alpha  - 2\cos \alpha  = 1\\{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\end{array} \right.\)

\( \Rightarrow {\left( {1 + 2\cos \alpha } \right)^2} + {\cos ^2}\alpha  = 1\)

\( \Rightarrow 5{\cos ^2}\alpha  + 4\cos \alpha  = 0 \Leftrightarrow \cos \alpha  =  - \frac{4}{5}\)

\( \Rightarrow \sin \alpha  =  - \sqrt {1 - {{\cos }^2}\alpha }  =  - \frac{3}{5}\)

\( \Rightarrow \tan \alpha  = \frac{3}{4},\cot \alpha  = \frac{4}{3}\).

Ta có: \(P = 2\tan \left( {\alpha  + 5\pi } \right) + \cot \left( {3\pi  - \alpha } \right) = 2\tan \alpha  - \cot \alpha  = 2.\frac{3}{4} - \frac{4}{3} = \frac{1}{6}\).