If `cos(theta-alpha) = a` and `cos(theta-beta) = b` then the value of `sin^2(alpha-beta) + 2ab cos(alpha-beta)`

Answer

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. **Express Sine and Cosine in Terms of a and b:**
From the given information, we can express \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \) as follows:
\[ \sin(\theta - \alpha) = \sqrt{1 - a^2} \]
\[ \sin(\theta - \beta) = \sqrt{1 - b^2} \]

2. **Find \( \sin(\alpha - \beta) \):**
We can express \( \sin(\alpha - \beta) \) using the sine subtraction formula:
\[ \sin(\alpha - \beta) = \sin(\theta - \alpha) \cos(\theta - \beta) - \cos(\theta - \alpha) \sin(\theta - \beta \]
Substituting the values we have:
\[ \sin(\alpha - \beta) = \sqrt{1 - a^2} \cdot b - a \cdot \sqrt{1 - b^2} \]

3. **Calculate \( \sin^2(\alpha - \beta) \):**
Now, we need to square the expression for \( \sin(\alpha - \beta) \):
\[ \sin^2(\alpha - \beta) = \left(\sqrt{1 - a^2} \cdot b - a \cdot \sqrt{1 - b^2}\right)^2 \]
Expanding this:
\[ = (1 - a^2)b^2 - 2ab\sqrt{1 - a^2}\sqrt{1 - b^2} + a^2(1 - b^2) \]
\[ = b^2 - a^2b^2 - 2ab\sqrt{(1 - a^2)(1 - b^2)} + a^2 - a^2b^2 \]
\[ = a^2 + b^2 - 2a^2b^2 - 2ab\sqrt{(1 - a^2)(1 - b^2)} \]

4. **Find \( \cos(\alpha - \beta) \):**
Using the cosine subtraction formula:
\[ \cos(\alpha - \beta) = \cos(\theta - \beta) \cos(\theta - \alpha) + \sin(\theta - \beta) \sin(\theta - \alpha) \]
Substituting the values:
\[ = b \cdot a + \sqrt{1 - b^2} \cdot \sqrt{1 - a^2} \]
\[ = ab + \sqrt{(1 - a^2)(1 - b^2)} \]

5. **Substitute into the Original Expression:**
Now, substitute \( \sin^2(\alpha - \beta) \) and \( \cos(\alpha - \beta) \) into the original expression:
\[ \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) = \left(a^2 + b^2 - 2a^2b^2 - 2ab\sqrt{(1 - a^2)(1 - b^2)}\right) + 2ab(ab + \sqrt{(1 - a^2)(1 - b^2)}) \]
Simplifying this gives:
\[ = a^2 + b^2 - 2a^2b^2 + 2a^2b^2 + 2ab\sqrt{(1 - a^2)(1 - b^2)} \]
\[ = a^2 + b^2 \]

### Final Answer:
Thus, the value of \( \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) \) is:
\[ \boxed{a^2 + b^2} \]