If `(sinA+sinB+sinC)^(2)=sin^(2)A+sin^(2)B+sin^(2)C`, then which one is true?

Options

A. `sinA+sinB+sinC=0`
B. `cosA+cosB+cosC=0`
C. `(1)/(sinA)+(1)/(sinB)+(1)/(sinC)=0`
D. none of these

Answer

To solve the equation \((\sin A + \sin B + \sin C)^2 = \sin^2 A + \sin^2 B + \sin^2 C\), we can follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side using the identity \((x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz)\).

\[ (\sin A + \sin B + \sin C)^2 = \sin^2 A + \sin^2 B + \sin^2 C + 2(\sin A \sin B + \sin B \sin C + \sin C \sin A) \]

### Step 2: Set the equation
Now we set the expanded left-hand side equal to the right-hand side:

\[ \sin^2 A + \sin^2 B + \sin^2 C + 2(\sin A \sin B + \sin B \sin C + \sin C \sin A) = \sin^2 A + \sin^2 B + \sin^2 C \]

### Step 3: Simplify the equation
Next, we can simplify the equation by subtracting \(\sin^2 A + \sin^2 B + \sin^2 C\) from both sides:

\[ 2(\sin A \sin B + \sin B \sin C + \sin C \sin A) = 0 \]

### Step 4: Divide by 2
Dividing both sides by 2 gives us:

\[ \sin A \sin B + \sin B \sin C + \sin C \sin A = 0 \]

### Step 5: Analyze the result
This equation implies that the sum of the products of the sine functions is zero. This can occur in several scenarios, such as when one or more of the sine terms are zero or when the angles \(A\), \(B\), and \(C\) are such that their sine values cancel each other out.

### Conclusion
From the analysis, we can conclude that the condition \(\sin A \sin B + \sin B \sin C + \sin C \sin A = 0\) must hold true.