Vectors `vec(A)` and `vec(B)` include an angle `theta` between them If `(vec(A)+vec(B))` and `(vec(A)-vec(B))` respectively subtend angles `apha` and `beta` with A, then `(tan alpha+tan beta)` is

Options

A. `((AB 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))`

Answer

To solve the problem, we need to find the expression for \( \tan \alpha + \tan \beta \) given the vectors \( \vec{A} \) and \( \vec{B} \) and the angles \( \alpha \) and \( \beta \) they subtend with \( \vec{A} \).

### Step 1: Understand the Geometry
We have two vectors \( \vec{A} \) and \( \vec{B} \) that form an angle \( \theta \) between them. We need to visualize the vectors \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \).

### Step 2: Draw the Vectors
- Draw vector \( \vec{A} \) horizontally.
- Draw vector \( \vec{B} \) at an angle \( \theta \) from \( \vec{A} \).
- The resultant vector \( \vec{A} + \vec{B} \) can be represented as the diagonal of the parallelogram formed by \( \vec{A} \) and \( \vec{B} \).
- The vector \( \vec{A} - \vec{B} \) can be represented similarly.

### Step 3: Find \( \tan \alpha \)
In the triangle formed by \( \vec{A} \), \( \vec{B} \), and \( \vec{A} + \vec{B} \):
- The height (perpendicular) from the tip of \( \vec{B} \) to the line of \( \vec{A} \) is \( B \sin \theta \).
- The base (horizontal component) is \( A + B \cos \theta \).

Thus,
\[ \tan \alpha = \frac{B \sin \theta}{A + B \cos \theta} \]

### Step 4: Find \( \tan \beta \)
In the triangle formed by \( \vec{A} \), \( -\vec{B} \), and \( \vec{A} - \vec{B} \):
- The height (perpendicular) from the tip of \( -\vec{B} \) to the line of \( \vec{A} \) is again \( B \sin \theta \).
- The base (horizontal component) is \( A - B \cos \theta \).

Thus,
\[ \tan \beta = \frac{B \sin \theta}{A - B \cos \theta} \]

### Step 5: Combine \( \tan \alpha \) and \( \tan \beta \)
Now, we need to find \( \tan \alpha + \tan \beta \):
\[ \tan \alpha + \tan \beta = \frac{B \sin \theta}{A + B \cos \theta} + \frac{B \sin \theta}{A - B \cos \theta} \]

### Step 6: Find a Common Denominator
The common denominator for the two fractions is:
\[ (A + B \cos \theta)(A - B \cos \theta) = A^2 - B^2 \cos^2 \theta \]

Thus,
\[ \tan \alpha + \tan \beta = \frac{B \sin \theta (A - B \cos \theta) + B \sin \theta (A + B \cos \theta)}{A^2 - B^2 \cos^2 \theta} \]

### Step 7: Simplify the Numerator
The numerator simplifies to:
\[ B \sin \theta (A - B \cos \theta + A + B \cos \theta) = 2AB \sin \theta \]

### Final Result
Thus, we have:
\[ \tan \alpha + \tan \beta = \frac{2AB \sin \theta}{A^2 - B^2 \cos^2 \theta} \]

### Summary
The final expression for \( \tan \alpha + \tan \beta \) is:
\[ \tan \alpha + \tan \beta = \frac{2AB \sin \theta}{A^2 - B^2 \cos^2 \theta} \]