If (1+sinx)(1+siny)(1+sinz)=(1-sinx)(1-siny)(1-sinz)=k, then k has the value

Options

A. `pm` cos x cos y cos z
B. `pm` sin x sin y sin z
C. `pm` 3 sin x sin y sin z
D. None of these

Answer

To solve the problem, we start with the given equations:

1. \((1 + \sin x)(1 + \sin y)(1 + \sin z) = k\)
2. \((1 - \sin x)(1 - \sin y)(1 - \sin z) = k\)

We need to find the value of \(k\).

### Step 1: Multiply the two equations
We can multiply both sides of the two equations together:

\[ (1 + \sin x)(1 + \sin y)(1 + \sin z) \cdot (1 - \sin x)(1 - \sin y)(1 - \sin z) = k \cdot k = k^2 \]

### Step 2: Apply the identity
Using the identity \( (a + b)(a - b) = a^2 - b^2 \), we can simplify the left side:

- For \(1 + \sin x\) and \(1 - \sin x\):
\[ (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x \]
- Similarly, we can apply this for \(y\) and \(z\):
\[ (1 + \sin y)(1 - \sin y) = 1 - \sin^2 y = \cos^2 y \]
\[ (1 + \sin z)(1 - \sin z) = 1 - \sin^2 z = \cos^2 z \]

### Step 3: Combine the results
Now we can combine these results:

\[ \cos^2 x \cdot \cos^2 y \cdot \cos^2 z = k^2 \]

### Step 4: Take the square root
Taking the square root of both sides gives us:

\[ k = \pm \cos x \cdot \cos y \cdot \cos z \]

### Final Result
Thus, the value of \(k\) is:

\[ k = \pm \cos x \cdot \cos y \cdot \cos z \]