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

Options

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

Answer

To solve the problem, we need to find the value of 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 know that the sum of the angles in a triangle is \( A + B + C = 180^\circ \).

### Step 1: Use the property of angles in a triangle
Since \( A + B + C = 180^\circ \), we can choose specific values for \( A, B, C \) to simplify our calculations. A common choice is to set \( A = B = C = 60^\circ \).

### Step 2: Calculate \( \sin^2 60^\circ \) and \( \cos 60^\circ \)
We know:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
Thus,
\[ \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

Also,
\[ \cos 60^\circ = \frac{1}{2} \]

### Step 3: Substitute values into the expression
Now substitute \( A = B = C = 60^\circ \) into the expression:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \sin^2 60^\circ + \sin^2 60^\circ + \sin^2 60^\circ = 3 \times \frac{3}{4} = \frac{9}{4} \]

Next, calculate \( 2 \cos A \cos B \cos C \):
\[ 2 \cos A \cos B \cos C = 2 \times \cos 60^\circ \times \cos 60^\circ \times \cos 60^\circ = 2 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = 2 \times \frac{1}{8} = \frac{1}{4} \]

### Step 4: Combine the results
Now, we can substitute these results back into the expression:
\[ \sin^2 A + \sin^2 B + \sin^2 C - 2 \cos A \cos B \cos C = \frac{9}{4} - \frac{1}{4} = \frac{8}{4} = 2 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{2} \]