In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$. What is $\tan A$?

From SOHCAHTOA, we conclude that sin A = BC/AC and cos A = AB/AC. We have BC/AC=2AB/AC, or BC=2AB. You want tan A, which is BC/AB by SOHCAHTOA. But you have BC=2AB, so BC/AB=2, which is your answer.