If `sin(theta+alpha)=a,cos^(2)(theta+beta)=b,` then` sin(alpha-beta)=`

Options

A. `ab-(a-a^(2))(1-b^(2))`
B. `ab-sqrt((1-a)^(2)(1-b))`
C. `+-{asqrtb-sqrt((1-a^(2))(1-b))}`
D. `bsqrta-sqrt((1-a)^(2)(1-b))`

Explanation

To solve the problem, we need to find the value of \( \sin(\alpha - \beta) \) given that \( \sin(\theta + \alpha) = a \) and \( \cos^2(\theta + \beta) = b \).

### Step-by-Step Solution:

1. **Use the given identities:**
We know that:
\[ \sin(\theta + \alpha) = a \]
and
\[ \cos^2(\theta + \beta) = b \]
From the second equation, we can express \( \cos(\theta + \beta) \):
\[ \cos(\theta + \beta) = \pm \sqrt{b} \]

2. **Express \( \sin(\alpha - \beta) \):**
We can use the sine subtraction formula:
\[ \sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \]

3. **Express \( \sin(\alpha) \) and \( \cos(\alpha) \):**
From \( \sin(\theta + \alpha) = a \), we can write:
\[ \sin(\alpha) = a \cos(\theta) + \cos(\alpha) \sin(\theta) \]
We can also find \( \cos(\alpha) \) using the Pythagorean identity:
\[ \cos^2(\alpha) = 1 - \sin^2(\alpha) \]

4. **Express \( \sin(\beta) \) and \( \cos(\beta) \):**
From \( \cos^2(\theta + \beta) = b \), we can express \( \sin(\beta) \):
\[ \sin(\beta) = \sqrt{1 - b} \]
And we already have \( \cos(\beta) = \pm \sqrt{b} \).

5. **Substituting back into the sine subtraction formula:**
Now we substitute \( \sin(\alpha) \) and \( \cos(\beta) \) into the sine subtraction formula:
\[ \sin(\alpha - \beta) = (a \cos(\theta) + \cos(\alpha) \sin(\theta)) \cdot \sqrt{b} - \cos(\alpha) \cdot \sqrt{1 - b} \]

6. **Final expression:**
After substituting and simplifying, we can express \( \sin(\alpha - \beta) \) in terms of \( a \) and \( b \):
\[ \sin(\alpha - \beta) = a \sqrt{b} - \cos(\alpha) \sqrt{1 - b} \]

### Conclusion:
The value of \( \sin(\alpha - \beta) \) can be expressed as:
\[ \sin(\alpha - \beta) = a \sqrt{b} - \sqrt{1 - a^2} \sqrt{1 - b} \]