Question

[Nhok_baobinh]

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\). lại có : \(\sin\alpha=\frac{2}{3}\)

=> \(\frac{4}{9}+\cos^2\alpha=1\)

=> \(\cos^2\alpha=\frac{5}{9}\Rightarrow\cos\alpha=\frac{\sqrt{5}}{3}\)

Mà \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}:\frac{\sqrt{5}}{3}=\frac{2}{\sqrt{5}}\)

mặt khác: \(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{\sqrt{5}}{2}\)