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

Answer

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

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

### Step 2: Square both sides
Square both sides of the equation:
\[ (a \cos \theta)^2 = (c - b \sin \theta)^2 \]
This results in:
\[ a^2 \cos^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

### Step 3: Substitute \( \cos^2 \theta \)
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), substitute into the equation:
\[ a^2 (1 - \sin^2 \theta) = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]
Expanding this gives:
\[ a^2 - a^2 \sin^2 \theta = c^2 - 2bc \sin \theta + b^2 \sin^2 \theta \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ a^2 - c^2 = (b^2 + a^2) \sin^2 \theta - 2bc \sin \theta \]

### Step 5: Form a quadratic equation
This can be rearranged into a standard quadratic form:
\[ (b^2 + a^2) \sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0 \]

### Step 6: Use the product of roots
For a quadratic equation \( Ax^2 + Bx + C = 0 \), the product of the roots \( r_1 \) and \( r_2 \) is given by \( \frac{C}{A} \). Here, \( A = b^2 + a^2 \), \( B = -2bc \), and \( C = c^2 - a^2 \). Thus, the product of the roots \( \sin \alpha \sin \beta \) is:
\[ \sin \alpha \sin \beta = \frac{c^2 - a^2}{b^2 + a^2} \]

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

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