If `(1 + sin alpha) (1 + sin beta) (1 + sin gamma) = (1 - sinalpha) ( 1 - sin beta) ( 1- sin gamma)` = `k`. find the value of k?

Options

A. `pm cos alpha cos beta cos gamma`
B. `pm sin alphasin beta sin gamma`
C. `pm sin alpha cos beta cos gamma`
D. '+ sin alpha sin beta cos gamma`

Answer

To solve the equation \((1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) = (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = k\), we will follow these steps:

### Step 1: Set Up the Equations
We have two products that are equal to \(k\):
\[ (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) = k \]
\[ (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = k \]

### Step 2: Multiply Both Sides
We can multiply both sides of the first equation by the second equation:
\[ (1 + \sin \alpha)(1 - \sin \alpha)(1 + \sin \beta)(1 - \sin \beta)(1 + \sin \gamma)(1 - \sin \gamma) = k^2 \]

### Step 3: Apply the Difference of Squares
Using the identity \( (a + b)(a - b) = a^2 - b^2 \), we can simplify each pair:
\[ (1 - \sin^2 \alpha)(1 - \sin^2 \beta)(1 - \sin^2 \gamma) = k^2 \]
Since \(1 - \sin^2 x = \cos^2 x\), we can rewrite this as:
\[ \cos^2 \alpha \cos^2 \beta \cos^2 \gamma = k^2 \]

### Step 4: Take the Square Root
Taking the square root of both sides gives us:
\[ k = \pm \cos \alpha \cos \beta \cos \gamma \]

### Step 5: Conclusion
Thus, the value of \(k\) can be expressed as:
\[ k = \pm \cos \alpha \cos \beta \cos \gamma \]