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

Options

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

Answer

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

### Step 1: Express \( \sin^2(\alpha - \beta) \)
Using the identity for sine, we can express \( \sin^2(\alpha - \beta) \) in terms of cosine:
\[ \sin^2(\alpha - \beta) = 1 - \cos^2(\alpha - \beta) \]

### Step 2: Express \( \cos(\alpha - \beta) \)
Using the cosine subtraction formula:
\[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \]

### Step 3: Find \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \)
From the given equations:
\[ \cos(\theta - \alpha) = a \implies \sin^2(\theta - \alpha) = 1 - a^2 \]
\[ \cos(\theta - \beta) = b \implies \sin^2(\theta - \beta) = 1 - b^2 \]

### Step 4: Use the sine subtraction formula
We can express \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \) as:
\[ \sin(\theta - \alpha) = \sqrt{1 - a^2}, \quad \sin(\theta - \beta) = \sqrt{1 - b^2} \]

### Step 5: Substitute into the cosine formula
Now substituting into the cosine formula:
\[ \cos(\alpha - \beta) = \cos(\theta - \beta)\cos(\theta - \alpha) + \sin(\theta - \beta)\sin(\theta - \alpha) \]
Substituting the values we have:
\[ \cos(\alpha - \beta) = b \cdot a + \sqrt{1 - b^2} \cdot \sqrt{1 - a^2} \]

### Step 6: Substitute back into the original expression
Now substituting \( \sin^2(\alpha - \beta) \) and \( 2ab \cos(\alpha - \beta) \) into the expression:
\[ \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) = (1 - \cos^2(\alpha - \beta)) + 2ab(b \cdot a + \sqrt{1 - b^2} \cdot \sqrt{1 - a^2}) \]

### Step 7: Simplify the expression
After substituting and simplifying, we will find that:
\[ \sin^2(\alpha - \beta) + 2ab \cos(\alpha - \beta) = a^2 + b^2 \]

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