Ta có: \(\tan\alpha=\frac{đối}{kề}\)
\(\frac{\sin\alpha}{\cos\alpha}=\frac{đối}{huyền}:\frac{kề}{huyền}=\frac{đối}{huyền}.\frac{huyền}{kề}=\frac{đối}{kề}\)
Vậy \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\frac{\sin a}{\cos a}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{BC}.\frac{BC}{AC}=\frac{AB}{AC}=\tan a\)
=> ĐPCM