a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2a=1-\sin^2\alpha=1-\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{4}\)
\(\Rightarrow\cos\alpha=\frac{1}{2}\)(do \(\cos\alpha>0\))
b) \(Q=\sin^2\alpha+\cot^2\alpha.\sin^2\alpha=\sin^2\alpha\left(1+\cot^2\alpha\right)\)\(=\sin^2\alpha\left(1+\frac{\cos^2\alpha}{\sin^2\alpha}\right)=\sin^2\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=1\)
a) \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)
