If `k=(1-sinalpha)(1-sinbeta)(1-singamma)=(1+sinalpha)(1+sinbeta)(1+singamma)`, then the value of k is:

Options

A. `+- sinalpha sinbeta sin gamma`
B. `+- cosalpha cosbeta cos gamma`
C. `+-secalpha secbeta secgamma`
D. `+-"cosec"alpha "cosec"beta "cosec"gamma`

Answer

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

### Step 1: Set up the equations
We have two expressions for \( k \):
\[ k = (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) \]
\[ k = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \]

### Step 2: Multiply both sides
Since both expressions are equal to \( k \), we can multiply them:
\[ (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \]

### Step 3: Expand both sides
Using the identity \( (a - b)(a + b) = a^2 - b^2 \), we can expand both sides:
\[ k^2 = (1 - \sin^2 \alpha)(1 - \sin^2 \beta)(1 - \sin^2 \gamma) \]
This simplifies to:
\[ k^2 = \cos^2 \alpha \cos^2 \beta \cos^2 \gamma \]

### Step 4: Take the square root
To find \( k \), we take the square root of both sides:
\[ k = \pm \cos \alpha \cos \beta \cos \gamma \]

### Conclusion
Thus, the value of \( k \) is:
\[ k = \pm \cos \alpha \cos \beta \cos \gamma \]