7. 已知 $\cos \left(\alpha-\frac{\pi}{3}\right)=\cos \alpha$, 则 $\tan \alpha=$
$\cos \left(\alpha-\frac{\pi}{3}\right)=\cos \alpha \cos \frac{\pi}{3}+\sin \alpha \sin \frac{\pi}{3}=\frac{1}{2} \cos \alpha+\frac{\sqrt{3}}{2} \sin \alpha=\cos \alpha ， \therefore \frac{\sqrt{3}}{2} \sin \alpha=\frac{1}{2} \cos \alpha$ ，

$\therefore \frac{\sin \alpha}{\cos \alpha}=\frac{\sqrt{3}}{3}$ ，即 $\tan \alpha=\frac{\sqrt{3}}{3}$.

答案: $\frac{\sqrt{3}}{3}$