In a right triangle ABC right-angled at B. if `t a n A=1`, then value of `2sinA cos A=.`

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To solve the problem, we need to find the value of \(2 \sin A \cos A\) given that \(\tan A = 1\) in a right triangle ABC, where angle B is the right angle.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
Since \(\tan A = 1\), we know that:
\[ \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \]
This implies that \(BC = AB\) because \(\tan A = 1\).

2. **Assigning Lengths**:
Let's assign lengths to the sides:
- Let \(AB = k\) (the length of the adjacent side)
- Let \(BC = k\) (the length of the opposite side)
Thus, both sides are equal.

3. **Finding the Hypotenuse**:
Using the Pythagorean theorem, we can find the hypotenuse \(AC\):
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{k^2 + k^2} = \sqrt{2k^2} = k\sqrt{2} \]

4. **Calculating \(\sin A\) and \(\cos A\)**:
Now, we can find \(\sin A\) and \(\cos A\):
- \(\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}\)
- \(\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}}\)

5. **Calculating \(2 \sin A \cos A\)**:
Now, we substitute \(\sin A\) and \(\cos A\) into the expression \(2 \sin A \cos A\):
\[ 2 \sin A \cos A = 2 \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = 2 \cdot \frac{1}{2} = 1 \]

### Conclusion:
Thus, we have shown that:
\[ 2 \sin A \cos A = 1 \]