`sin alpha+sin beta=(1)/(4)` and `cos alpha+cos beta=(1)/(3)`
The value of `cos(alpha+beta)` is

Options

A. `(12)/(25)`
B. `(7)/(25)`
C. `(12)/(13)`
D. none of these.

Explanation

To find the value of \( \cos(\alpha + \beta) \) given the equations \( \sin \alpha + \sin \beta = \frac{1}{4} \) and \( \cos \alpha + \cos \beta = \frac{1}{3} \), we can follow these steps:

### Step-by-Step Solution

1. **Use the Sum-to-Product Identities:**
We can use the sum-to-product identities for sine and cosine:
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
From the problem, we have:
\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{4} \quad \text{(1)} \]
\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{3} \quad \text{(2)} \]

2. **Solve for \( \cos\left(\frac{\alpha - \beta}{2}\right) \):**
From equations (1) and (2), we can express \( \cos\left(\frac{\alpha - \beta}{2}\right) \):
\[ \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{8 \sin\left(\frac{\alpha + \beta}{2}\right)} \quad \text{(from (1))} \]
\[ \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{6 \cos\left(\frac{\alpha + \beta}{2}\right)} \quad \text{(from (2))} \]

3. **Equate the Two Expressions:**
Setting the two expressions for \( \cos\left(\frac{\alpha - \beta}{2}\right) \) equal to each other:
\[ \frac{1}{8 \sin\left(\frac{\alpha + \beta}{2}\right)} = \frac{1}{6 \cos\left(\frac{\alpha + \beta}{2}\right)} \]
Cross-multiplying gives:
\[ 6 = 8 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha + \beta}{2}\right) \]
Using the identity \( \sin(2x) = 2 \sin(x) \cos(x) \):
\[ 6 = 4 \sin\left(\alpha + \beta\right) \]
Thus,
\[ \sin\left(\alpha + \beta\right) = \frac{3}{2} \]
This is incorrect since the sine function cannot exceed 1. We need to re-evaluate our equations.

4. **Revisit the Equations:**
From the equations we derived:
\[ \sin\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{8 \cos\left(\frac{\alpha - \beta}{2}\right)} \]
\[ \cos\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{6 \cos\left(\frac{\alpha - \beta}{2}\right)} \]
We can substitute \( x = \cos\left(\frac{\alpha - \beta}{2}\right) \):
\[ \sin\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{8x}, \quad \cos\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{6x} \]
Then using \( \sin^2 + \cos^2 = 1 \):
\[ \left(\frac{1}{8x}\right)^2 + \left(\frac{1}{6x}\right)^2 = 1 \]
Simplifying gives:
\[ \frac{1}{64x^2} + \frac{1}{36x^2} = 1 \]
Finding a common denominator:
\[ \frac{36 + 64}{2304x^2} = 1 \implies 100 = 2304x^2 \implies x^2 = \frac{100}{2304} \implies x = \frac{10}{48} = \frac{5}{24} \]

5. **Find \( \cos(\alpha + \beta) \):**
Now we can find \( \cos(\alpha + \beta) \) using:
\[ \cos^2\left(\frac{\alpha + \beta}{2}\right) + \sin^2\left(\frac{\alpha + \beta}{2}\right) = 1 \]
Thus:
\[ \cos(\alpha + \beta) = 1 - \sin^2(\alpha + \beta) \]
Using the sine value derived:
\[ \cos(\alpha + \beta) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]

### Final Answer
The value of \( \cos(\alpha + \beta) \) is \( \frac{7}{25} \).