From algebra we know that if `ax^(2) +bx + c=0 , a( ne 0), b, c int R` has roots `alpha` and `beta` then `alpha + beta=-b/a` and `alpha beta = c/a`. Trignometric functions `sin theta` and `cos theta, tan theta` and `sec theta, " cosec " theta` and `cot theta` obey `sin^(2)theta + cos^(2)theta =1`. A linear relation in `sin theta` and `cos theta, sec theta` and `tan theta` or `" cosec "theta` and `cos theta` can be transformed into a quadratic equation in, say, `sin theta, tan theta` or `cot theta` respectively. And then one can apply sum and product of roots to find the desired result. Let `a cos theta, b sintheta=c` have two roots `theta_(1)` and `theta_(2)`. `theta_(1) ne theta_(2)`.
The vlaue of `cos(theta_(1) + theta_(2))` is a and b not being simultaneously zero)

Options

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

Answer

To solve the problem, we need to find the value of \( \cos(\theta_1 + \theta_2) \) given the equation \( a \cos \theta + b \sin \theta = c \), where \( \theta_1 \) and \( \theta_2 \) are the roots of this equation.

### 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-hand side:
\[ a^2 \cos^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

4. **Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \)**:
Substitute \( \sin^2 \theta \):
\[ a^2 \cos^2 \theta = c^2 - 2bc \sin \theta + b^2 (1 - \cos^2 \theta) \]
Simplifying gives:
\[ a^2 \cos^2 \theta = c^2 - 2bc \sin \theta + b^2 - b^2 \cos^2 \theta \]

5. **Combine like terms**:
\[ (a^2 + b^2) \cos^2 \theta + 2bc \sin \theta - c^2 - b^2 = 0 \]

6. **Identify the quadratic in terms of \( \sin \theta \)**:
This equation can be treated as a quadratic equation in \( \sin \theta \):
\[ b^2 \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \]

7. **Apply the sum and product of roots**:
Let \( \sin \theta_1 \) and \( \sin \theta_2 \) be the roots. From the quadratic formula, we know:
- Sum of roots \( \sin \theta_1 + \sin \theta_2 = \frac{2bc}{b^2} \)
- Product of roots \( \sin \theta_1 \sin \theta_2 = \frac{c^2 - a^2}{b^2} \)

8. **Find \( \cos(\theta_1 + \theta_2) \)**:
Using the cosine addition formula:
\[ \cos(\theta_1 + \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \]
We can express \( \cos \theta_1 \cos \theta_2 \) in terms of \( \sin \theta_1 \) and \( \sin \theta_2 \):
\[ \cos \theta_1 \cos \theta_2 = \sqrt{1 - \sin^2 \theta_1} \sqrt{1 - \sin^2 \theta_2} \]

9. **Substituting values**:
Substitute the values of \( \sin \theta_1 + \sin \theta_2 \) and \( \sin \theta_1 \sin \theta_2 \) into the cosine addition formula to find \( \cos(\theta_1 + \theta_2) \).

10. **Final expression**:
After simplifying, we find:
\[ \cos(\theta_1 + \theta_2) = \frac{a^2 - b^2}{a^2 + b^2} \]

### Conclusion:
Thus, the value of \( \cos(\theta_1 + \theta_2) \) is given by:
\[ \cos(\theta_1 + \theta_2) = \frac{a^2 - b^2}{a^2 + b^2} \]