Okay, so I need to find tan A in a right triangle ABC where angle B is 90 degrees, and sin A is equal to 2 cos A. Hmm, let's start by recalling some basic trigonometric identities and relationships in right triangles. 

First, in a right-angled triangle, the sum of the other two angles is 90 degrees. So, angle A and angle C must add up to 90 degrees. That might come in handy later. Also, since we're dealing with sin, cos, and tan of angle A, I should remember that tan A is sin A divided by cos A. The problem states that sin A is twice cos A, which might allow me to express sin A in terms of cos A or vice versa and then use the Pythagorean identity to find their exact values.

Wait, let's write down what we know. The problem says sin A = 2 cos A. So, if I denote sin A as 2x and cos A as x, then according to the Pythagorean identity, sin²A + cos²A = 1. Let me substitute those expressions into the identity. That would give (2x)² + x² = 1. Simplifying that, that's 4x² + x² = 1, which is 5x² = 1. Then x² = 1/5, so x = sqrt(1/5) or x = -sqrt(1/5). But since we're dealing with a right triangle, all trigonometric function values should be positive. So, x = 1/sqrt(5). Therefore, sin A = 2/sqrt(5) and cos A = 1/sqrt(5). 

Therefore, tan A, which is sin A divided by cos A, would be (2/sqrt(5)) / (1/sqrt(5)) = 2. So, tan A is 2. Wait, that seems straightforward. Let me check if I did that correctly. 

Alternatively, maybe I can approach this using triangle sides. In triangle ABC, angle B is 90 degrees, so sides opposite angles A, B, and C are a, b, c respectively, where c is the hypotenuse. Then, sin A would be opposite over hypotenuse, which is a/c, and cos A would be adjacent over hypotenuse, which is b/c. The problem states that sin A = 2 cos A, so a/c = 2*(b/c). Multiplying both sides by c gives a = 2b. So, the side opposite angle A is twice the length of the side adjacent to angle A. 

Then, using Pythagoras, a² + b² = c². Since a = 2b, substituting gives (2b)² + b² = c² => 4b² + b² = 5b² = c². So, c = b*sqrt(5). Therefore, the sides are in the ratio 2b : b : b*sqrt(5). Then, tan A is opposite over adjacent, which is a/b = 2b/b = 2. So, same result. 

Wait, so both methods give tan A = 2. That seems consistent. Let me verify once more. If sin A = 2 cos A, then tan A = sin A / cos A = 2. So, yes, that's a direct conclusion. But why did we go through the Pythagorean identity earlier? Because if we didn't know that tan A is sin A over cos A, we might have had to compute sin and cos separately. But since the problem directly relates sin and cos, dividing them gives tan directly. So, in this case, tan A = 2. 

Therefore, after verifying through two different approaches—using trigonometric identities and using the side ratios of the right triangle—it's clear that tan A is 2. I don't see any flaws in this reasoning. Both methods lead to the same answer, so I'm confident that's correct.
Therefore, the value of $\tan A$ is $\boxed{2}$.
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