`(sin2A+sin2B+sin2C)/(sinA+sinB +sinC)` is equal to

Options

A. `8 sin((A)/(2))sin((B)/(2))sin((C)/(2))`
B. `8cos((A)/(2))cos((B)/(2))cos((C)/(2))`
C. `8tan((A)/(2))tan((B)/(2))tan((C)/(2))`
D. `8cot((A)/(2))cot((B)/(2))cot((C)/(2))`

Explanation

To solve the expression \((\sin 2A + \sin 2B + \sin 2C) / (\sin A + \sin B + \sin C)\), we can use trigonometric identities.

### Step-by-Step Solution:

1. **Use the Double Angle Identity**:
We know that \(\sin 2A = 2 \sin A \cos A\). Therefore, we can express \(\sin 2A + \sin 2B + \sin 2C\) as:
\[ \sin 2A + \sin 2B + \sin 2C = 2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C \]

2. **Factor Out the 2**:
We can factor out the 2 from the expression:
\[ = 2(\sin A \cos A + \sin B \cos B + \sin C \cos C) \]

3. **Use the Sum-to-Product Identities**:
We can apply the sum-to-product identities to simplify \(\sin A + \sin B + \sin C\). However, for this case, we will keep it as is for now.

4. **Substituting Back**:
Now we rewrite the original expression:
\[ \frac{2(\sin A \cos A + \sin B \cos B + \sin C \cos C)}{\sin A + \sin B + \sin C} \]

5. **Simplifying the Expression**:
The expression can be simplified further depending on the values of A, B, and C. However, we can use a known identity:
\[ \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]
This leads us to:
\[ \frac{4 \sin A \sin B \sin C}{\sin A + \sin B + \sin C} \]

6. **Final Result**:
Thus, the expression simplifies to:
\[ \frac{4 \sin A \sin B \sin C}{\sin A + \sin B + \sin C} \]

### Conclusion:
The expression \(\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C}\) is equal to \(\frac{4 \sin A \sin B \sin C}{\sin A + \sin B + \sin C}\).