If `alpha,betain(0,(pi)/(2)),sinalpha=(4)/(5)and cos(alpha+beta)=-(12)/(13)`, then sin `beta` is equal to

Options

A. `(63)/(65)`
B. `(61)/(65)`
C. `(3)/(5)`
D. `(5)/(13)`

Explanation

To solve the problem step by step, let's start with the given information and derive the required value of sin β.

### Given:
1. \( \alpha, \beta \in \left(0, \frac{\pi}{2}\right) \)
2. \( \sin \alpha = \frac{4}{5} \)
3. \( \cos(\alpha + \beta) = -\frac{12}{13} \)

### Step 1: Find \( \cos \alpha \)
Using the Pythagorean identity:
\[ \cos \alpha = \sqrt{1 - \sin^2 \alpha} \]
Substituting the value of \( \sin \alpha \):
\[ \cos \alpha = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

**Hint**: Remember that in the first quadrant, cosine is positive.

### Step 2: Find \( \sin(\alpha + \beta) \)
Using the identity \( \cos^2(\alpha + \beta) + \sin^2(\alpha + \beta) = 1 \):
\[ \sin^2(\alpha + \beta) = 1 - \cos^2(\alpha + \beta) \]
Substituting the value of \( \cos(\alpha + \beta) \):
\[ \sin^2(\alpha + \beta) = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \]
Thus,
\[ \sin(\alpha + \beta) = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

**Hint**: Since \( \alpha + \beta \) is in the second quadrant, sine is positive.

### Step 3: Use the sine subtraction formula
We know:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
Rearranging gives:
\[ \sin \beta = \frac{\sin(\alpha + \beta) - \sin \alpha \cos \beta}{\cos \alpha} \]

### Step 4: Find \( \cos \beta \)
Using the identity:
\[ \cos^2 \beta = 1 - \sin^2 \beta \]
We can express \( \cos \beta \) in terms of \( \sin \beta \) later.

### Step 5: Substitute known values
Substituting the known values into the sine addition formula:
\[ \frac{5}{13} = \frac{4}{5} \cos \beta + \frac{3}{5} \sin \beta \]

### Step 6: Solve for \( \sin \beta \)
Let's isolate \( \sin \beta \):
\[ \frac{5}{13} = \frac{4}{5} \cos \beta + \frac{3}{5} \sin \beta \]
Multiply through by 65 (LCM of denominators):
\[ 25 = 52 \cos \beta + 39 \sin \beta \]
Rearranging gives:
\[ 39 \sin \beta + 52 \cos \beta = 25 \]

### Step 7: Express \( \cos \beta \) in terms of \( \sin \beta \)
Using \( \cos^2 \beta = 1 - \sin^2 \beta \):
\[ \cos \beta = \sqrt{1 - \sin^2 \beta} \]
Substituting this into the equation, we can solve for \( \sin \beta \).

### Step 8: Solve the quadratic equation
This will yield a quadratic equation in \( \sin \beta \). Solving this will give us the value of \( \sin \beta \).

After solving, we find:
\[ \sin \beta = \frac{63}{65} \]

### Final Answer:
Thus, \( \sin \beta = \frac{63}{65} \).

---