a) Ta có: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=3\Rightarrow sin\alpha=3cos\alpha\)
Thế vào đề \(\Rightarrow\dfrac{cos\alpha+3cos\alpha}{cos\alpha-3cos\alpha}=\dfrac{4cos\alpha}{-2cos\alpha}=-2\)
b) Ta có: \(sin^3\alpha+cos^3\alpha=\left(sin\alpha+cos\alpha\right)\left(sin^2\alpha+cos^2-sin\alpha.cos\alpha\right)\)
\(=\left(sin\alpha+cos\alpha\right)\left(1-0,48\right)=\dfrac{13}{25}\left(sin\alpha+cos\alpha\right)\)
Ta có \(\left(sin\alpha+cos\alpha\right)^2=sin^2\alpha+cos^2+2.sin\alpha.cos\alpha=1+2.0,48=\dfrac{49}{25}\)
\(\Rightarrow sin\alpha+cos\alpha=\dfrac{7}{5}\Rightarrow\dfrac{13}{25}\left(sin\alpha+cos\alpha\right)=\dfrac{13}{25}.\dfrac{7}{5}=\dfrac{91}{125}\)