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

Answer

To find the value of \( \cos \alpha \cos \beta \) where \( \alpha \) and \( \beta \) are the solutions of the equation \( a \cos \theta + b \sin \theta = c \), we can follow these steps:

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

### Step 2: Squaring Both Sides
Square both sides of the equation:
\[ (c - a \cos \theta)^2 = (b \sin \theta)^2 \]

### Step 3: Expanding Both Sides
Expanding both sides results in:
\[ c^2 - 2ac \cos \theta + a^2 \cos^2 \theta = b^2 \sin^2 \theta \]

### Step 4: Substituting \( \sin^2 \theta \)
Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ c^2 - 2ac \cos \theta + a^2 \cos^2 \theta = b^2 (1 - \cos^2 \theta) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ c^2 - 2ac \cos \theta + a^2 \cos^2 \theta = b^2 - b^2 \cos^2 \theta \]
Combine like terms:
\[ (a^2 + b^2) \cos^2 \theta - 2ac \cos \theta + (c^2 - b^2) = 0 \]

### Step 6: Identifying Coefficients
This is a quadratic equation in terms of \( \cos \theta \):
\[ A \cos^2 \theta + B \cos \theta + C = 0 \]
Where:
- \( A = a^2 + b^2 \)
- \( B = -2ac \)
- \( C = c^2 - b^2 \)

### Step 7: Product of Roots
The product of the roots \( \cos \alpha \cos \beta \) for a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ \cos \alpha \cos \beta = \frac{C}{A} \]

### Step 8: Substituting Values
Substituting the values of \( C \) and \( A \):
\[ \cos \alpha \cos \beta = \frac{c^2 - b^2}{a^2 + b^2} \]

### Final Answer
Thus, the value of \( \cos \alpha \cos \beta \) is:
\[ \cos \alpha \cos \beta = \frac{c^2 - b^2}{a^2 + b^2} \]

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