ta có : \(tan\alpha+cot\alpha=3\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}=3\)
\(\Leftrightarrow\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}=3\Leftrightarrow\dfrac{1}{sin\alpha.cos\alpha}=3\)
\(\Leftrightarrow sin\alpha.cos\alpha=\dfrac{1}{3}\) vậy \(sin\alpha.cos\alpha=\dfrac{1}{3}\)