If `cos alpha + cosbeta=a, sinalpha + sinbeta=b and theta ` is the arithmetic mean between `alpha` and `beta` , then `(sin2theta + cos 2theta)` is equal to :

Options

A. `((a+b)^(2))/(a^2 + b^2)`
B. `((a-b)^(2))/(a^2 + b^2)`
C. `(a^(2)-b^(2))/(a^2 + b^2)`
D. None of these

Answer

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

### Step-by-Step Solution

1. **Express \( \theta \)**:
\[ \theta = \frac{\alpha + \beta}{2} \]

2. **Use the sum-to-product identities**:
From the given equations, we can express \( \cos \alpha + \cos \beta \) and \( \sin \alpha + \sin \beta \) using the sum-to-product identities:
\[ \cos \alpha + \cos \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]

Substituting \( \theta \) into these identities gives:
\[ a = 2 \cos(\theta) \cos\left(\frac{\alpha - \beta}{2}\right) \]
\[ b = 2 \sin(\theta) \cos\left(\frac{\alpha - \beta}{2}\right) \]

3. **Divide the equations**:
Dividing the two equations:
\[ \frac{b}{a} = \frac{2 \sin(\theta) \cos\left(\frac{\alpha - \beta}{2}\right)}{2 \cos(\theta) \cos\left(\frac{\alpha - \beta}{2}\right)} \]
Simplifying gives:
\[ \frac{b}{a} = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta) \]

4. **Construct a right triangle**:
Let’s consider a right triangle where:
- Opposite side = \( b \)
- Adjacent side = \( a \)
- Hypotenuse = \( \sqrt{a^2 + b^2} \)

5. **Find \( \sin \theta \) and \( \cos \theta \)**:
From the triangle:
\[ \sin(\theta) = \frac{b}{\sqrt{a^2 + b^2}}, \quad \cos(\theta) = \frac{a}{\sqrt{a^2 + b^2}} \]

6. **Calculate \( \sin 2\theta \) and \( \cos 2\theta \)**:
Using the double angle formulas:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
\[ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \]

Substituting the values:
\[ \sin(2\theta) = 2 \left(\frac{b}{\sqrt{a^2 + b^2}}\right) \left(\frac{a}{\sqrt{a^2 + b^2}}\right) = \frac{2ab}{a^2 + b^2} \]

For \( \cos(2\theta) \):
\[ \cos(2\theta) = \left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 - \left(\frac{b}{\sqrt{a^2 + b^2}}\right)^2 = \frac{a^2 - b^2}{a^2 + b^2} \]

7. **Combine \( \sin 2\theta \) and \( \cos 2\theta \)**:
Now, we can find \( \sin(2\theta) + \cos(2\theta) \):
\[ \sin(2\theta) + \cos(2\theta) = \frac{2ab}{a^2 + b^2} + \frac{a^2 - b^2}{a^2 + b^2} \]
\[ = \frac{2ab + a^2 - b^2}{a^2 + b^2} \]

### Final Result
Thus, the expression \( \sin 2\theta + \cos 2\theta \) is:
\[ \sin 2\theta + \cos 2\theta = \frac{a^2 + 2ab - b^2}{a^2 + b^2} \]