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

Options

A. `(24)/(25)`
B. `(13)/(25)`
C. `(12)/(13)`
D. none of these.

Explanation

To solve the problem, we are given the equations:

1. \( \sin \alpha + \sin \beta = \frac{1}{4} \)
2. \( \cos \alpha + \cos \beta = \frac{1}{3} \)

We need to find the value of \( \sin(\alpha + \beta) \).

### Step-by-Step Solution:

**Step 1: Use the sum-to-product identities.**

We can express \( \sin \alpha + \sin \beta \) and \( \cos \alpha + \cos \beta \) using the sum-to-product identities:

\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]

\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]

**Step 2: Set up the equations.**

From the first equation:

\[ 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{4} \]

From the second equation:

\[ 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{3} \]

**Step 3: Simplify the equations.**

Dividing both equations by 2 gives us:

\[ \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{8} \quad \text{(Equation 1)} \]

\[ \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) = \frac{1}{6} \quad \text{(Equation 2)} \]

**Step 4: Divide Equation 1 by Equation 2.**

Now, we can divide Equation 1 by Equation 2:

\[ \frac{\sin\left(\frac{\alpha + \beta}{2}\right)}{\cos\left(\frac{\alpha + \beta}{2}\right)} = \frac{\frac{1}{8}}{\frac{1}{6}} \]

This simplifies to:

\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{1}{8} \cdot \frac{6}{1} = \frac{3}{4} \]

**Step 5: Find \( \sin(\alpha + \beta) \).**

Using the double angle formula for sine:

\[ \sin(2\theta) = \frac{2 \tan(\theta)}{1 + \tan^2(\theta)} \]

Let \( \theta = \frac{\alpha + \beta}{2} \). Then:

\[ \sin(\alpha + \beta) = \sin(2\theta) = \frac{2 \tan\left(\frac{\alpha + \beta}{2}\right)}{1 + \tan^2\left(\frac{\alpha + \beta}{2}\right)} \]

Substituting \( \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{3}{4} \):

\[ \sin(\alpha + \beta) = \frac{2 \cdot \frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} \]

Calculating \( \left(\frac{3}{4}\right)^2 = \frac{9}{16} \):

\[ \sin(\alpha + \beta) = \frac{\frac{6}{4}}{1 + \frac{9}{16}} = \frac{\frac{6}{4}}{\frac{16 + 9}{16}} = \frac{\frac{6}{4}}{\frac{25}{16}} = \frac{6 \cdot 16}{4 \cdot 25} = \frac{96}{100} = \frac{24}{25} \]

### Final Answer:

Thus, the value of \( \sin(\alpha + \beta) \) is:

\[ \boxed{\frac{24}{25}} \]