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

Options

A. `ab+sqrt(1-a^(2))sqrt(1-b^(2))`
B. `ab-sqrt(1-a^(2))sqrt(1-b^(2))`
C. `asqrt(1-b^(2))-bsqrt(1-a^(2))`
D. `asqrt(1-b^(2))+bsqrt(1-a^(2))`

Explanation

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

### Step-by-step Solution:

1. **Write down the given equations:**
\[ \cos(\theta - \alpha) = a \]
\[ \cos(\theta - \beta) = b \]

2. **Express \( \sin(\theta - \alpha) \) and \( \sin(\theta - \beta) \):**
Using the Pythagorean identity, we can find the sine values:
\[ \sin(\theta - \alpha) = \sqrt{1 - a^2} \]
\[ \sin(\theta - \beta) = \sqrt{1 - b^2} \]

3. **Use the cosine subtraction formula:**
The cosine of the difference of two angles can be expressed as:
\[ \cos(\alpha - \beta) = \cos((\theta - \alpha) + (\theta - \beta)) = \cos(\theta - \alpha)\cos(\theta - \beta) + \sin(\theta - \alpha)\sin(\theta - \beta) \]

4. **Substitute the known values:**
Plugging in the values we have:
\[ \cos(\alpha - \beta) = a \cdot b + \sqrt{1 - a^2} \cdot \sqrt{1 - b^2} \]

5. **Simplify the expression:**
The final expression for \( \cos(\alpha - \beta) \) becomes:
\[ \cos(\alpha - \beta) = ab + \sqrt{(1 - a^2)(1 - b^2)} \]

### Final Answer:
\[ \cos(\alpha - \beta) = ab + \sqrt{(1 - a^2)(1 - b^2)} \]