If A, B and C are the angles of a triangle and
`|(1,1,1),(1 + sin A,1 + sin B,1 + sin C),(sin A + sin^(2) A,sin B + sin^(2)B,sin C + sin^(2) C)|= 0`, then the triangle ABC is

Options

A. equilateral
B. isosceles
C. any triangle
D. right angled

Answer

To solve the given determinant problem, we need to analyze the determinant and simplify it step by step. The determinant is given as:

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

We are given that \(D = 0\).

### Step 1: Simplify the Determinant
We will perform row operations to simplify the determinant. First, we subtract the first row from the second row:

\[ R_2 \rightarrow R_2 - R_1 \]

This gives us:

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

### Step 2: Further Simplify
Next, we will perform another row operation by subtracting the second row from the third row:

\[ R_3 \rightarrow R_3 - R_2 \]

This results in:

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

### Step 3: Column Operations
Now, we will perform column operations. We will subtract the second column from the first column and the third column from the second column:

\[ C_1 \rightarrow C_1 - C_2, \quad C_2 \rightarrow C_2 - C_3 \]

This gives us:

\[ D = \begin{vmatrix} 0 & 1 - \sin B & 1 - \sin C \\ \sin A - \sin B & \sin B - \sin C & \sin C \\ \sin^2 A - \sin^2 B & \sin^2 B - \sin^2 C & \sin^2 C \end{vmatrix} \]

### Step 4: Expand the Determinant
Now we will expand the determinant along the first row:

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

### Step 5: Factor the Determinant
Notice that \(\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B)\) and \(\sin^2 B - \sin^2 C = (\sin B - \sin C)(\sin B + \sin C)\). Thus, we can factor out:

\[ D = (\sin A - \sin B)(\sin B - \sin C)(\sin A + \sin B + \sin C) = 0 \]

### Step 6: Analyze the Factors
For the determinant to be zero, at least one of the factors must be zero:

1. \(\sin A - \sin B = 0 \Rightarrow A = B\)
2. \(\sin B - \sin C = 0 \Rightarrow B = C\)
3. \(\sin C - \sin A = 0 \Rightarrow C = A\)

If any two angles are equal, the triangle is isosceles.

### Conclusion
Since we have shown that at least two angles of triangle ABC are equal, we conclude that triangle ABC is an **isosceles triangle**.