If in a triangle ABC, `|{:(1, sin A ,sin^(2)A),(1,sin B , sin^(2)B),(1, sin C, sin^(2)C):}|= 0` then the triangle is

Options

A. equilateral or isosceles
B. equilateral or right-angled
C. right angled or isosceles
D. None ot these

Answer

To solve the problem, we need to evaluate the determinant given in the question and analyze the conditions under which it equals zero. The determinant is given as:

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

### Step 1: Evaluate the Determinant
We start by applying properties of determinants. We can subtract the first row from the second and third rows:

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

### Step 2: Factor the Rows
Next, we can factor the differences of squares in the second and third rows:

\[ \sin^2 B - \sin^2 A = (\sin B - \sin A)(\sin B + \sin A) \]
\[ \sin^2 C - \sin^2 A = (\sin C - \sin A)(\sin C + \sin A) \]

Thus, we rewrite the determinant as:

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

### Step 3: Simplify the Determinant
Now, we can factor out \((\sin B - \sin A)\) from the second row and \((\sin C - \sin A)\) from the third row:

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

### Step 4: Evaluate the Remaining Determinant
The remaining determinant simplifies to:

\[ D = (\sin B - \sin A)(\sin C - \sin A) \begin{vmatrix} 1 & \sin A \\ 1 & \sin B + \sin A \\ 1 & \sin C + \sin A \end{vmatrix} \]

This determinant equals zero if either \((\sin B - \sin A) = 0\) or \((\sin C - \sin A) = 0\).

### Step 5: Analyze the Conditions
1. If \(\sin B = \sin A\), then \(B = A\) or \(B + A = \pi\). Since \(A + B + C = \pi\), the second condition is not possible in a triangle.
2. If \(\sin C = \sin A\), then \(C = A\) or \(C + A = \pi\). Again, the second condition is not possible in a triangle.
3. Similarly, if \(\sin C = \sin B\), then \(C = B\) or \(C + B = \pi\).

### Conclusion
From these conditions, we conclude that:
- If any two angles are equal (e.g., \(A = B\), \(B = C\), or \(C = A\)), the triangle is isosceles.
- If all three angles are equal, the triangle is equilateral.

Thus, the triangle can either be isosceles or equilateral.

### Final Answer
The triangle is either **isosceles or equilateral**.

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