Question

helppp

Answer 1

\(cos\alpha=\dfrac{3}{5}=>cos^2\alpha=\dfrac{9}{25}\)

có \(sin^2\alpha+cos^2\alpha=1< =>sin^2\alpha+\dfrac{9}{25}=1=>sin\alpha=\dfrac{4}{5}\left(sin\alpha>0\right)\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\)

\(cot\alpha\cdot tan\alpha=1=>cot\alpha\cdot\dfrac{4}{3}=1=>cot\alpha=\dfrac{3}{4}\)