Vectors `vecA` and `vecB` include an angle `theta` between them. If `(vecA + vecB)` and `(vecA- vecB)` respectively subtend angles `alpha and beta` with `vecA`, then `(tan alpha + tan beta)` is

Options

A. `((A sin theta))/((A^(2) +B^(2)cos^(2)theta)) `
B. `((2AB sin theta))/((A^(2) -B^(2)cos^(2)theta)) `
C. `((A^(2) sin^(2) theta))/((A^(2) +B^(2)cos^(2)theta)) `
D. `((B^(2) sin^(2) theta))/((A^(2) -B^(2)cos^(2)theta)) `

Explanation

To solve the problem, we need to find the value of \( \tan \alpha + \tan \beta \) given the vectors \( \vec{A} \) and \( \vec{B} \) with an angle \( \theta \) between them. The angles \( \alpha \) and \( \beta \) are formed by the vectors \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) with respect to \( \vec{A} \).

### Step-by-Step Solution:

1. **Understanding the Angles**:
- The angle \( \alpha \) is the angle between \( \vec{A} \) and \( \vec{A} + \vec{B} \).
- The angle \( \beta \) is the angle between \( \vec{A} \) and \( \vec{A} - \vec{B} \).

2. **Using the Definition of Tangent**:
- For the angle \( \alpha \):
\[ \tan \alpha = \frac{|\vec{B}| \sin \theta}{|\vec{A}| + |\vec{B}| \cos \theta} \]
- For the angle \( \beta \):
\[ \tan \beta = \frac{|\vec{B}| \sin \theta}{|\vec{A}| - |\vec{B}| \cos \theta} \]

3. **Adding the Two Tangents**:
- Now we need to find \( \tan \alpha + \tan \beta \):
\[ \tan \alpha + \tan \beta = \frac{|\vec{B}| \sin \theta}{|\vec{A}| + |\vec{B}| \cos \theta} + \frac{|\vec{B}| \sin \theta}{|\vec{A}| - |\vec{B}| \cos \theta} \]

4. **Finding a Common Denominator**:
- The common denominator for the two fractions is:
\[ (|\vec{A}| + |\vec{B}| \cos \theta)(|\vec{A}| - |\vec{B}| \cos \theta) \]
- This can be simplified using the difference of squares:
\[ |\vec{A}|^2 - |\vec{B}|^2 \cos^2 \theta \]

5. **Combining the Numerators**:
- The numerator becomes:
\[ |\vec{B}| \sin \theta (|\vec{A}| - |\vec{B}| \cos \theta + |\vec{A}| + |\vec{B}| \cos \theta) = |\vec{B}| \sin \theta (2 |\vec{A}|) \]

6. **Final Expression**:
- Thus, we have:
\[ \tan \alpha + \tan \beta = \frac{2 |\vec{A}| |\vec{B}| \sin \theta}{|\vec{A}|^2 - |\vec{B}|^2 \cos^2 \theta} \]

### Conclusion:
The final result for \( \tan \alpha + \tan \beta \) is:
\[ \tan \alpha + \tan \beta = \frac{2 |\vec{A}| |\vec{B}| \sin \theta}{|\vec{A}|^2 - |\vec{B}|^2 \cos^2 \theta} \]