In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$.  What is $\tan A$?
To find $\tan A$, I need to use the trigonometric identity $\tan A = \frac{\sin A}{\cos A}$. I already have an equation that relates $\sin A$ and $\cos A$, so I can substitute it into the identity and get $\tan A = \frac{2\cos A}{\cos A} = 2$. To do that, I can use the Pythagorean theorem, which states that $\sin^2 A + \cos^2 A = 1$ for any angle $A$ in a right triangle. Plugging in $\sin A = 2\cos A$, I get $(2\cos A)^2 + \cos^2 A = 1$, which simplifies to $5\cos^2 A = 1$, or $\cos A = \pm \frac{1}{\sqrt{5}}$. Since $A$ is an acute angle, I can only take the positive value of $\cos A$, so $\cos A = \frac{1}{\sqrt{5}}$ and $\sin A = \frac{2}{\sqrt{5}}$. These values agree with my earlier answer of $\tan A = 2$, because $\frac{2}{\sqrt{5}} \div \frac{1}{\sqrt{5}} = 2$. Therefore, I can conclude that $\tan A = 2$ is the only possible answer for this problem. 