If A,B,C are the angles of a triangle, then `sin^(2) A + sin^(2) B + sin^(2) C - 2 cos A cos B cos C=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the expression \( \sin^2 A + \sin^2 B + \sin^2 C - 2 \cos A \cos B \cos C \) where \( A, B, C \) are the angles of a triangle, we can follow these steps:

### Step 1: Use the identity for sine squared
We start by rewriting \( \sin^2 A \), \( \sin^2 B \), and \( \sin^2 C \) using the identity:
\[ \sin^2 A = 1 - \cos^2 A \]
Thus, we can express the sum as:
\[ \sin^2 A + \sin^2 B + \sin^2 C = (1 - \cos^2 A) + (1 - \cos^2 B) + (1 - \cos^2 C) = 3 - (\cos^2 A + \cos^2 B + \cos^2 C) \]

### Step 2: Substitute into the original expression
Now, substituting this back into the original expression gives us:
\[ 3 - (\cos^2 A + \cos^2 B + \cos^2 C) - 2 \cos A \cos B \cos C \]

### Step 3: Use the identity for cosine squared
Next, we can use the identity for \( \cos^2 A + \cos^2 B + \cos^2 C \). For angles of a triangle, we have:
\[ \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C \]
Thus, we can substitute this into our expression:
\[ 3 - (1 - 2 \cos A \cos B \cos C) - 2 \cos A \cos B \cos C \]

### Step 4: Simplify the expression
Now simplifying this gives:
\[ 3 - 1 + 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C = 2 \]

### Final Result
Thus, we find that:
\[ \sin^2 A + \sin^2 B + \sin^2 C - 2 \cos A \cos B \cos C = 2 \]