It is given that `cos(theta-alpha)=a, cos(theta-beta)=b`
What is `sin^(2)(alpha-beta)+2abcos(alpha-beta)` equal to ?

Options

A. `a^(2)+b^(2)`
B. `a^(2)-b^(2)`
C. `b^(2)-a^(2)`
D. `-(a^(2)+b^(2))`

Explanation

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

### Step-by-Step Solution:

1. **Start with the given expression:**
\[ x = \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) \]

2. **Use the identity for \( \sin^2(\alpha - \beta) \):**
\[ \sin^2(\alpha - \beta) = 1 - \cos^2(\alpha - \beta) \]
Therefore, we can rewrite \( x \):
\[ x = 1 - \cos^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) \]

3. **Let \( \cos(\alpha - \beta) = c \):**
Now substituting \( c \) into the expression:
\[ x = 1 - c^2 + 2abc \]

4. **Rearranging the expression:**
\[ x = 1 + 2abc - c^2 \]

5. **Use the cosine subtraction formula:**
We know:
\[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \]
From the given conditions, we have:
\[ \cos(\alpha) = a \quad \text{and} \quad \cos(\beta) = b \]
Thus, we can express \( \sin(\alpha) \) and \( \sin(\beta) \):
\[ \sin(\alpha) = \sqrt{1 - a^2} \quad \text{and} \quad \sin(\beta) = \sqrt{1 - b^2} \]

6. **Substituting back:**
Now substituting back into the cosine subtraction formula:
\[ c = ab + \sqrt{1 - a^2}\sqrt{1 - b^2} \]

7. **Substituting \( c \) into the expression for \( x \):**
\[ x = 1 + 2ab(ab + \sqrt{1 - a^2}\sqrt{1 - b^2}) - (ab + \sqrt{1 - a^2}\sqrt{1 - b^2})^2 \]

8. **Simplifying the expression:**
This will lead us to:
\[ x = 1 - (1 - a^2)(1 - b^2) + a^2 + b^2 \]

9. **Final simplification:**
After simplifying, we find:
\[ x = a^2 + b^2 \]

### Conclusion:
Thus, the required value is:
\[ \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) = a^2 + b^2 \]