If a,b,and c are the side of a triangle and A,B and C are the angles opposite to a,b,and c respectively, then
`Delta=|{:(a^(2),bsinA,CsinA),(bsinA,1,cosA),(CsinA,cos A,1):}|` is independent of

Options

A. (A)a
B. (B)b=0
C. (C)c
D. (D)A,B,C

Answer

To solve the problem, we need to analyze the determinant given in the question and show that it is independent of the sides \( a, b, c \) of the triangle.

The determinant is given as:

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

### Step 1: Use the Sine Law
Using the sine law, we know that:

\[ \sin A = \frac{a}{k}, \quad \sin B = \frac{b}{k}, \quad \sin C = \frac{c}{k} \]

where \( k \) is a constant related to the circumradius of the triangle.

### Step 2: Substitute the Sine Values
Substituting the sine values into the determinant, we have:

\[ \Delta = \begin{vmatrix} a^2 & b \frac{a}{k} & c \frac{a}{k} \\ b \frac{a}{k} & 1 & \cos A \\ c \frac{a}{k} & \cos A & 1 \end{vmatrix} \]

### Step 3: Factor out \( a \)
We can factor out \( a \) from the first column:

\[ \Delta = a^2 \begin{vmatrix} 1 & b \frac{1}{k} & c \frac{1}{k} \\ b \frac{1}{k} & 1 & \cos A \\ c \frac{1}{k} & \cos A & 1 \end{vmatrix} \]

### Step 4: Simplify the Determinant
Now, we can simplify the determinant further. Let’s denote the new determinant as \( D \):

\[ D = \begin{vmatrix} 1 & \frac{b}{k} & \frac{c}{k} \\ \frac{b}{k} & 1 & \cos A \\ \frac{c}{k} & \cos A & 1 \end{vmatrix} \]

### Step 5: Apply Column Operations
Next, we can apply column operations to simplify \( D \). We will perform the following operations:
- Column 2 = Column 2 - \( \frac{b}{k} \) * Column 1
- Column 3 = Column 3 - \( \frac{c}{k} \) * Column 1

This will yield:

\[ D = \begin{vmatrix} 1 & 0 & 0 \\ \frac{b}{k} & 1 - \left(\frac{b}{k}\right)^2 & \cos A - \left(\frac{b}{k}\right) \\ \frac{c}{k} & \cos A - \left(\frac{c}{k}\right) & 1 - \left(\frac{c}{k}\right)^2 \end{vmatrix} \]

### Step 6: Calculate the Determinant
Now, we can calculate the determinant \( D \). The first row simplifies the determinant calculation, and we can expand it:

\[ D = 1 \cdot \begin{vmatrix} 1 - \left(\frac{b}{k}\right)^2 & \cos A - \left(\frac{b}{k}\right) \\ \cos A - \left(\frac{c}{k}\right) & 1 - \left(\frac{c}{k}\right)^2 \end{vmatrix} \]

### Step 7: Final Result
After calculating the determinant, we find that \( D \) is a function of angles \( A, B, C \) and does not depend on the sides \( a, b, c \). Therefore, we conclude that:

\[ \Delta = a^2 \cdot D \]

Since \( D \) is independent of \( a, b, c \), we can say that \( \Delta \) is also independent of the sides of the triangle.

### Conclusion
Thus, we can conclude that \( \Delta \) is independent of the sides \( a, b, c \) of the triangle.