If `( 1+ sin A ) ( 1 + sin B ) ( 1 + sin C ) = ( 1 - sin A ) ( 1- sin B ) ( 1- sin C )`, then each side is equal to

Options

A. `+-` sin A sin B sinC
B. `+-` cos A cos B cos C
C. `+-` sin A cos B cos C
D. `+-` cos A sin B sin C

Answer

To solve the equation \( (1 + \sin A)(1 + \sin B)(1 + \sin C) = (1 - \sin A)(1 - \sin B)(1 - \sin C) \), we will follow these steps:

### Step 1: Define the expression
Let us define the left-hand side and right-hand side of the equation as \( k \):
\[ k = (1 + \sin A)(1 + \sin B)(1 + \sin C) = (1 - \sin A)(1 - \sin B)(1 - \sin C) \]

### Step 2: Square both sides
To eliminate the products, we can square both sides:
\[ k^2 = \left((1 + \sin A)(1 + \sin B)(1 + \sin C)\right) \left((1 - \sin A)(1 - \sin B)(1 - \sin C)\right) \]

### Step 3: Simplify the products
Using the identity \( (1 + x)(1 - x) = 1 - x^2 \), we can simplify:
\[ k^2 = (1 + \sin A)(1 - \sin A) \cdot (1 + \sin B)(1 - \sin B) \cdot (1 + \sin C)(1 - \sin C) \]
This becomes:
\[ k^2 = (1 - \sin^2 A)(1 - \sin^2 B)(1 - \sin^2 C) \]

### Step 4: Use the Pythagorean identity
Recall that \( 1 - \sin^2 x = \cos^2 x \). Therefore, we can rewrite the expression:
\[ k^2 = \cos^2 A \cdot \cos^2 B \cdot \cos^2 C \]

### Step 5: Take the square root
Taking the square root of both sides gives us:
\[ k = \pm \cos A \cdot \cos B \cdot \cos C \]

### Conclusion
Thus, we find that:
\[ (1 + \sin A)(1 + \sin B)(1 + \sin C) = (1 - \sin A)(1 - \sin B)(1 - \sin C) = \pm \cos A \cdot \cos B \cdot \cos C \]