In a `triangleABC`, right-angled at B, if AB : AC = 1:2, then find the value of
`(2tanA)/(1+tan^2A)`.

Answer

To solve the problem step by step, we will follow the given information and apply trigonometric identities.

### Step 1: Understand the triangle
We have triangle ABC which is right-angled at B. We are given that the ratio of sides AB to AC is 1:2.

### Step 2: Assign lengths based on the ratio
Let AB = x and AC = 2x. Since AC is the hypotenuse, we can denote:
- AB (opposite side to angle A) = x
- BC (adjacent side to angle A) = y
- AC (hypotenuse) = 2x

### Step 3: Apply the Pythagorean theorem
Using the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values we have:
\[ (2x)^2 = x^2 + y^2 \]
\[ 4x^2 = x^2 + y^2 \]
\[ 4x^2 - x^2 = y^2 \]
\[ 3x^2 = y^2 \]
\[ y = \sqrt{3}x \]

### Step 4: Find tan A
Now, we can find the value of \( \tan A \):
\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}} \]

### Step 5: Substitute tan A into the expression
We need to find the value of:
\[ \frac{2 \tan A}{1 + \tan^2 A} \]
Substituting \( \tan A = \frac{1}{\sqrt{3}} \):
\[ \tan^2 A = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
Now substituting into the expression:
\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} \]

### Step 6: Simplify the denominator
The denominator simplifies as follows:
\[ 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \]

### Step 7: Complete the expression
Now substituting back:
\[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \]

### Step 8: Rationalize the denominator
To rationalize the denominator:
\[ \frac{3}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the final value of \( \frac{2 \tan A}{1 + \tan^2 A} \) is:
\[ \frac{\sqrt{3}}{2} \]

---