If ` alpha ` and ` beta ` are 2 distinct roots of equation ` a cos theta + b sin theta = C ` then ` cos( alpha + beta ) = `

Options

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

Answer

To solve the problem, we need to find the value of \( \cos(\alpha + \beta) \) given that \( \alpha \) and \( \beta \) are distinct roots of the equation \( a \cos \theta + b \sin \theta = C \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a \cos \theta + b \sin \theta = C \]

2. **Rearrange the equation:**
\[ a \cos \theta = C - b \sin \theta \]

3. **Square both sides:**
\[ a^2 \cos^2 \theta = (C - b \sin \theta)^2 \]
Expanding the right side gives:
\[ a^2 \cos^2 \theta = C^2 - 2bC \sin \theta + b^2 \sin^2 \theta \]

4. **Substitute \( \cos^2 \theta = 1 - \sin^2 \theta \):**
\[ a^2 (1 - \sin^2 \theta) = C^2 - 2bC \sin \theta + b^2 \sin^2 \theta \]
This simplifies to:
\[ a^2 - a^2 \sin^2 \theta = C^2 - 2bC \sin \theta + b^2 \sin^2 \theta \]

5. **Rearranging gives:**
\[ (a^2 + b^2) \sin^2 \theta - 2bC \sin \theta + (C^2 - a^2) = 0 \]
This is a quadratic equation in \( \sin \theta \).

6. **Identify the coefficients:**
Let \( P = a^2 + b^2 \), \( Q = -2bC \), and \( R = C^2 - a^2 \).

7. **Use the product of roots formula:**
The product of the roots \( \sin \alpha \sin \beta \) is given by:
\[ \sin \alpha \sin \beta = \frac{R}{P} = \frac{C^2 - a^2}{a^2 + b^2} \]

8. **Now, find \( \cos \alpha \cos \beta \):**
From the original equation, rearranging gives:
\[ b \sin \theta = C - a \cos \theta \]
Squaring both sides gives:
\[ b^2 \sin^2 \theta = (C - a \cos \theta)^2 \]
Expanding this yields:
\[ b^2 \sin^2 \theta = C^2 - 2aC \cos \theta + a^2 \cos^2 \theta \]

9. **Substituting \( \sin^2 \theta = 1 - \cos^2 \theta \):**
\[ b^2 (1 - \cos^2 \theta) = C^2 - 2aC \cos \theta + a^2 \cos^2 \theta \]
Rearranging gives:
\[ (a^2 + b^2) \cos^2 \theta - 2aC \cos \theta + (C^2 - b^2) = 0 \]

10. **Again, use the product of roots formula:**
The product of the roots \( \cos \alpha \cos \beta \) is given by:
\[ \cos \alpha \cos \beta = \frac{C^2 - b^2}{a^2 + b^2} \]

11. **Now, apply the cosine addition formula:**
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

12. **Substituting the values:**
\[ \cos(\alpha + \beta) = \frac{C^2 - b^2}{a^2 + b^2} - \frac{C^2 - a^2}{a^2 + b^2} \]

13. **Combine the fractions:**
\[ \cos(\alpha + \beta) = \frac{(C^2 - b^2) - (C^2 - a^2)}{a^2 + b^2} = \frac{a^2 - b^2}{a^2 + b^2} \]

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