In a `Delta ABC`, a, b, c are sides and A, B, C are angles opposite to them, then the value of the determinant
`|(a^(2),b sin A,c sin A),(b sin A,1,cos A),(c sin A,cos A,1)|`, is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} a^2 & b \sin A & c \sin A \\ b \sin A & 1 & \cos A \\ c \sin A & \cos A & 1 \end{vmatrix} \]

we will use the properties of determinants to simplify and calculate it step by step.

### Step 1: Expand the Determinant

We can expand the determinant using the first row:

\[ D = a^2 \begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} - b \sin A \begin{vmatrix} b \sin A & \cos A \\ c \sin A & 1 \end{vmatrix} + c \sin A \begin{vmatrix} b \sin A & 1 \\ c \sin A & \cos A \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

1. **First 2x2 determinant:**

\[ \begin{vmatrix} 1 & \cos A \\ \cos A & 1 \end{vmatrix} = (1)(1) - (\cos A)(\cos A) = 1 - \cos^2 A = \sin^2 A \]

2. **Second 2x2 determinant:**

\[ \begin{vmatrix} b \sin A & \cos A \\ c \sin A & 1 \end{vmatrix} = (b \sin A)(1) - (\cos A)(c \sin A) = b \sin A - c \sin A \cos A \]

3. **Third 2x2 determinant:**

\[ \begin{vmatrix} b \sin A & 1 \\ c \sin A & \cos A \end{vmatrix} = (b \sin A)(\cos A) - (1)(c \sin A) = b \sin A \cos A - c \sin A \]

### Step 3: Substitute Back into the Determinant

Now substituting these values back into the expression for \(D\):

\[ D = a^2 \sin^2 A - b \sin A (b \sin A - c \sin A \cos A) + c \sin A (b \sin A \cos A - c \sin A) \]

### Step 4: Simplify the Expression

Expanding the terms:

1. The first term remains \(a^2 \sin^2 A\).
2. The second term becomes:

\[ -b \sin A (b \sin A) + b c \sin^2 A \cos A = -b^2 \sin^2 A + b c \sin^2 A \cos A \]

3. The third term becomes:

\[ c \sin A (b \sin A \cos A) - c^2 \sin A = b c \sin^2 A \cos A - c^2 \sin A \]

Combining all these:

\[ D = a^2 \sin^2 A - b^2 \sin^2 A + 2bc \sin^2 A \cos A - c^2 \sin A \]

### Step 5: Factor Out Common Terms

Notice that we can factor out \(\sin^2 A\):

\[ D = \sin^2 A (a^2 - b^2 + 2bc \cos A - c^2) \]

### Step 6: Use the Cosine Rule

Using the cosine rule in triangle \(ABC\):

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substituting this into our expression gives:

\[ D = \sin^2 A \left(a^2 - b^2 - c^2 + 2bc \left(\frac{b^2 + c^2 - a^2}{2bc}\right)\right) \]

This simplifies to:

\[ D = \sin^2 A \cdot 0 = 0 \]

### Final Result

Thus, the value of the determinant is:

\[ \boxed{0} \]