In triangle ABC, angle C is 90 degrees, cosA = √5 / 5. Find tgA.

1. By definition tgA = SinA / СosA.

2. Let us square both sides of the equality. We get: (tgA) ^ 2 = (SinA) ^ 2 / (CosA) ^ 2.

3. It is known that (SinA) ^ 2 = 1 – (CosA) ^ 2.

4. Substitute the expression for SinA into the equality of item 2. We get:
(tgA) ^ 2 = 1 / (CosA) ^ 2 – 1.

5. Substitute the value for CosA = √5 / 5 = 1 / √5. We get: (tgA) ^ 2 = 5 – 1 = 4.

6. Take the square root of both sides of the equality. We get: tgA = 2 and tgA = -2. The negative root does not correspond to the condition of the problem, since angle A is an acute angle of a right-angled triangle. That is, A <90 deg. Hence tgA> 0.

Answer: tgA = 2.