If `A+B+C=pi`, then the value of determinant `|{:(sin^(2)A, cotA,1),(sin^(2)B,cotB,1),(sin^(2)C,cot C,1):}|` is equal to

Options

A. 0
B. 1
C. `-1`
D. None of these

Answer

To solve the determinant

\[ D = \begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix} \]

given that \( A + B + C = \pi \), we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix} \]

### Step 2: Apply Row Operations
We can perform row operations to simplify the determinant. We can subtract the third column from the first two columns:

1. \( R_1 \rightarrow R_1 - R_3 \)
2. \( R_2 \rightarrow R_2 - R_3 \)
3. \( R_3 \) remains unchanged.

This gives us:

\[ D = \begin{vmatrix} \sin^2 A - 1 & \cot A - 1 & 1 \\ \sin^2 B - 1 & \cot B - 1 & 1 \\ \sin^2 C - 1 & \cot C - 1 & 1 \end{vmatrix} \]

### Step 3: Recognize Trigonometric Identities
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can rewrite \( \sin^2 A - 1 \) as \( -\cos^2 A \), \( \sin^2 B - 1 \) as \( -\cos^2 B \), and \( \sin^2 C - 1 \) as \( -\cos^2 C \). Therefore, we have:

\[ D = \begin{vmatrix} -\cos^2 A & \cot A - 1 & 1 \\ -\cos^2 B & \cot B - 1 & 1 \\ -\cos^2 C & \cot C - 1 & 1 \end{vmatrix} \]

### Step 4: Factor Out -1
We can factor out -1 from the first column:

\[ D = -1 \begin{vmatrix} \cos^2 A & \cot A - 1 & 1 \\ \cos^2 B & \cot B - 1 & 1 \\ \cos^2 C & \cot C - 1 & 1 \end{vmatrix} \]

### Step 5: Further Simplification
Now we can simplify the determinant further. Notice that \( \cot A = \frac{\cos A}{\sin A} \), so \( \cot A - 1 = \frac{\cos A - \sin A}{\sin A} \). Thus, we can rewrite the determinant as:

\[ D = -1 \begin{vmatrix} \cos^2 A & \frac{\cos A - \sin A}{\sin A} & 1 \\ \cos^2 B & \frac{\cos B - \sin B}{\sin B} & 1 \\ \cos^2 C & \frac{\cos C - \sin C}{\sin C} & 1 \end{vmatrix} \]

### Step 6: Evaluate the Determinant
Using properties of determinants and the fact that \( A + B + C = \pi \), we can conclude that the determinant evaluates to zero because the rows become linearly dependent due to the angle sum identity.

Thus, we find that:

\[ D = 0 \]

### Final Answer
The value of the determinant is:

\[ \boxed{0} \]