What is the value of `sin(A+B)sin(A-B)+sin(B+C)sin(B-C)+sin(C+A)sin(C-A)`?

Options

A. 0
B. `sinA+sinB+sinC`
C. `cosA+cosB+cosC`
D. 1

Explanation

To find the value of the expression \( \sin(A+B)\sin(A-B) + \sin(B+C)\sin(B-C) + \sin(C+A)\sin(C-A) \), we can use the product-to-sum identities in trigonometry.

### Step-by-Step Solution:

1. **Apply the Product-to-Sum Identity**:
The product-to-sum identity states that:
\[ \sin x \sin y = \frac{1}{2} [\cos(x-y) - \cos(x+y)] \]
We will apply this identity to each term in the expression.

2. **First Term**:
For \( \sin(A+B)\sin(A-B) \):
\[ \sin(A+B)\sin(A-B) = \frac{1}{2} [\cos((A+B)-(A-B)) - \cos((A+B)+(A-B))] \]
Simplifying this gives:
\[ = \frac{1}{2} [\cos(2B) - \cos(2A)] \]

3. **Second Term**:
For \( \sin(B+C)\sin(B-C) \):
\[ \sin(B+C)\sin(B-C) = \frac{1}{2} [\cos((B+C)-(B-C)) - \cos((B+C)+(B-C))] \]
Simplifying this gives:
\[ = \frac{1}{2} [\cos(2C) - \cos(2B)] \]

4. **Third Term**:
For \( \sin(C+A)\sin(C-A) \):
\[ \sin(C+A)\sin(C-A) = \frac{1}{2} [\cos((C+A)-(C-A)) - \cos((C+A)+(C-A))] \]
Simplifying this gives:
\[ = \frac{1}{2} [\cos(2A) - \cos(2C)] \]

5. **Combine All Terms**:
Now, we combine all three results:
\[ \sin(A+B)\sin(A-B) + \sin(B+C)\sin(B-C) + \sin(C+A)\sin(C-A) \]
becomes:
\[ = \frac{1}{2} [\cos(2B) - \cos(2A)] + \frac{1}{2} [\cos(2C) - \cos(2B)] + \frac{1}{2} [\cos(2A) - \cos(2C)] \]

6. **Simplifying the Expression**:
Combine the terms:
\[ = \frac{1}{2} \left( \cos(2B) - \cos(2A) + \cos(2C) - \cos(2B) + \cos(2A) - \cos(2C) \right) \]
Notice that \( \cos(2B) \) cancels out, \( \cos(2A) \) cancels out, and \( \cos(2C) \) cancels out:
\[ = \frac{1}{2} (0) = 0 \]

### Final Result:
Thus, the value of the expression is:
\[ \boxed{0} \]