Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=3\)
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3\dfrac{1}{3}-1}{2\dfrac{1}{3}+1}=\dfrac{2}{\dfrac{2}{3}+1}=\dfrac{6}{5}\)
b. \(B=\dfrac{sin^2\alpha-3sin\alpha cos\alpha+2}{2sin^2\alpha+sin\alpha cos\alpha+cos^2\alpha}\)
Chia tử và mẫu cho \(cos^2\alpha\) ta được:
\(B=\dfrac{\left(\dfrac{1}{3}\right)^2-3.\dfrac{1}{3}+\dfrac{2}{cos^2\alpha}}{2\left(\dfrac{1}{3}\right)^2+\dfrac{1}{3}+1}\)
Mà \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=3\) \(\Rightarrow cos\alpha=3sin\alpha\Leftrightarrow cos^2\alpha=9sin^2\alpha\)
Lại có: \(cos^2\alpha+sin^2\alpha=1\Leftrightarrow10sin^2\alpha=1\Rightarrow cos^2\alpha=1-sin^2\alpha=\dfrac{9}{10}\)
\(B=\dfrac{\left(\dfrac{1}{3}\right)^2-3.\dfrac{1}{3}+\dfrac{2}{\dfrac{9}{10}}}{2\left(\dfrac{1}{3}\right)^2+\dfrac{1}{3}+1}=\dfrac{6}{7}\)
