If tan `(alpha + beta)` = 2 and tan `(alpha - beta)` = 1, then tan (2`alpha`) is equal to:

Options

A. `-3`
B. `-2`
C. `-1/3`
D. 1

Answer

To solve the problem, we need to find the value of \( \tan(2\alpha) \) given that \( \tan(\alpha + \beta) = 2 \) and \( \tan(\alpha - \beta) = 1 \).

### Step-by-Step Solution:

1. **Use the formula for \( \tan(2\alpha) \)**:
We know that:
\[ \tan(2\alpha) = \tan((\alpha + \beta) + (\alpha - \beta)) \]
This can be expressed using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Here, let \( A = \alpha + \beta \) and \( B = \alpha - \beta \).

2. **Substitute the values of \( \tan A \) and \( \tan B \)**:
From the problem, we have:
\[ \tan(\alpha + \beta) = 2 \quad \text{and} \quad \tan(\alpha - \beta) = 1 \]
Therefore, substituting these values into the formula gives:
\[ \tan(2\alpha) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} = \frac{2 + 1}{1 - (2 \cdot 1)} \]

3. **Simplify the expression**:
Now, simplify the numerator and denominator:
\[ \tan(2\alpha) = \frac{3}{1 - 2} = \frac{3}{-1} = -3 \]

4. **Final result**:
Thus, we find that:
\[ \tan(2\alpha) = -3 \]

### Conclusion:
The value of \( \tan(2\alpha) \) is \( -3 \).

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