If `alpha` and `beta` are the solutions of a `cos theta + b sin theta =c`, then
`cos alpha +cos beta ="……………"`

Answer

To solve the problem, we need to find the expression for \( \cos \alpha + \cos \beta \) given that \( \alpha \) and \( \beta \) are the solutions of the equation \( a \cos \theta + b \sin \theta = c \).

### Step-by-Step Solution:

1. **Rearranging the Equation:**
Start with the given equation:
\[ a \cos \theta + b \sin \theta = c \]
Rearranging gives:
\[ c - a \cos \theta = b \sin \theta \]

2. **Squaring Both Sides:**
Square both sides to eliminate the sine and cosine terms:
\[ (c - a \cos \theta)^2 = (b \sin \theta)^2 \]

3. **Expanding Both Sides:**
Expanding the left side using the formula \( (x - y)^2 = x^2 - 2xy + y^2 \):
\[ c^2 - 2ac \cos \theta + a^2 \cos^2 \theta = b^2 \sin^2 \theta \]

4. **Substituting \( \sin^2 \theta \):**
Replace \( \sin^2 \theta \) with \( 1 - \cos^2 \theta \):
\[ c^2 - 2ac \cos \theta + a^2 \cos^2 \theta = b^2 (1 - \cos^2 \theta) \]

5. **Rearranging the Equation:**
This simplifies to:
\[ c^2 - 2ac \cos \theta + (a^2 + b^2) \cos^2 \theta - b^2 = 0 \]
Rearranging gives:
\[ (a^2 + b^2) \cos^2 \theta - 2ac \cos \theta + (c^2 - b^2) = 0 \]

6. **Identifying the Quadratic Form:**
This is a quadratic equation in terms of \( \cos \theta \):
\[ A \cos^2 \theta + B \cos \theta + C = 0 \]
where \( A = a^2 + b^2 \), \( B = -2ac \), and \( C = c^2 - b^2 \).

7. **Using the Sum of Roots Formula:**
The sum of the roots of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ \cos \alpha + \cos \beta = -\frac{B}{A} \]
Substituting the values of \( A \) and \( B \):
\[ \cos \alpha + \cos \beta = -\left(-\frac{2ac}{a^2 + b^2}\right) = \frac{2ac}{a^2 + b^2} \]

### Final Result:
Thus, we have:
\[ \cos \alpha + \cos \beta = \frac{2ac}{a^2 + b^2} \]