If ` cot ^(2) A - 3 =0 ` find :
sin 2A

Answer

To solve the equation \( \cot^2 A - 3 = 0 \) and find \( \sin 2A \), we can follow these steps:

### Step 1: Solve for \( \cot^2 A \)

Starting with the equation:
\[ \cot^2 A - 3 = 0 \]
We can rearrange this to find \( \cot^2 A \):
\[ \cot^2 A = 3 \]

### Step 2: Find \( \cot A \)

Taking the square root of both sides, we find:
\[ \cot A = \pm \sqrt{3} \]

### Step 3: Determine the angles for \( A \)

1. **For \( \cot A = \sqrt{3} \)**:
- The angle \( A \) that satisfies this is:
\[ A = \frac{\pi}{6} \quad \text{(in radians)} \]

2. **For \( \cot A = -\sqrt{3} \)**:
- The angle \( A \) that satisfies this is:
\[ A = -\frac{\pi}{6} \quad \text{(in radians)} \]

### Step 4: Find \( \sin 2A \)

Using the double angle formula \( \sin 2A = 2 \sin A \cos A \), we need to find \( \sin A \) and \( \cos A \) for both angles.

1. **For \( A = \frac{\pi}{6} \)**:
- \( \sin A = \frac{1}{2} \)
- \( \cos A = \frac{\sqrt{3}}{2} \)
- Therefore:
\[ \sin 2A = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

2. **For \( A = -\frac{\pi}{6} \)**:
- \( \sin A = -\frac{1}{2} \)
- \( \cos A = \frac{\sqrt{3}}{2} \)
- Therefore:
\[ \sin 2A = 2 \cdot -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \]

### Final Result

Thus, the two possible values for \( \sin 2A \) are:
\[ \sin 2A = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin 2A = -\frac{\sqrt{3}}{2} \]