If a,b,c are the sides of a `DeltaABC and A,B,C` are respectively the angles opposite to them , then
`|(a^2,bsinA, csinA),(bsinA,1,cos(B-C)),(csinA,cos(B-C),1)|` is equal to

Options

A. `sinA-sinB sin C`
B. abc
C. 1
D. 0

Answer

To solve the determinant

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

we will follow these steps:

### Step 1: Factor out common terms
We notice that \( \sin A \) is common in the first row and the first column. We can factor out \( a^2 \) from the first row and \( b \sin A \) from the second row, and \( c \sin A \) from the third row.

\[ D = a^2 \cdot \sin A \cdot \begin{vmatrix} 1 & 1 & 1 \\ 1 & \frac{1}{b \sin A} & \frac{\cos(B - C)}{b \sin A} \\ 1 & \frac{\cos(B - C)}{c \sin A} & \frac{1}{c \sin A} \end{vmatrix} \]

### Step 2: Simplify the determinant
Now, we can simplify the determinant. Let's denote the new determinant as \( D' \):

\[ D' = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \frac{1}{b \sin A} & \frac{\cos(B - C)}{b \sin A} \\ 1 & \frac{\cos(B - C)}{c \sin A} & \frac{1}{c \sin A} \end{vmatrix} \]

### Step 3: Apply row operations
We can perform row operations to simplify the determinant. Subtract the first row from the second and third rows:

\[ D' = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \frac{1}{b \sin A} - 1 & \frac{\cos(B - C)}{b \sin A} - 1 \\ 0 & \frac{\cos(B - C)}{c \sin A} - 1 & \frac{1}{c \sin A} - 1 \end{vmatrix} \]

### Step 4: Calculate the determinant
Now we can calculate the determinant \( D' \). The first column has become zero, which means that the determinant will evaluate to zero.

### Conclusion
Thus, the value of the original determinant \( D \) is:

\[ D = 0 \]