The value of the determinant `Delta=|{: ( sin^(2)23^(@)" "sin^(2)67^(@)" "cos180^(@)),(-sin^(2)67^(@)" "-sin^(2)23^(@)" "-cos180^(@)),(cos180^(@)" "sin^(2)23^(@)" "sin^(2)67^(@)):}|` is

Options

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

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} \sin^2 23^\circ & \sin^2 67^\circ & \cos 180^\circ \\ -\sin^2 67^\circ & -\sin^2 23^\circ & -\cos 180^\circ \\ \cos 180^\circ & \sin^2 23^\circ & \sin^2 67^\circ \end{vmatrix} \]

we will follow these steps:

### Step 1: Simplify the trigonometric values
We know that:
- \(\cos 180^\circ = -1\)
- \(\sin^2 67^\circ = \cos^2 23^\circ\) (using the identity \(\sin(90^\circ - \theta) = \cos \theta\))

Thus, we can rewrite the determinant as:

\[ \Delta = \begin{vmatrix} \sin^2 23^\circ & \cos^2 23^\circ & -1 \\ -\cos^2 23^\circ & -\sin^2 23^\circ & 1 \\ -1 & \sin^2 23^\circ & \cos^2 23^\circ \end{vmatrix} \]

### Step 2: Perform column operations
We will perform the operation \(C_1 \to C_1 + C_2\):

\[ C_1 = \begin{pmatrix} \sin^2 23^\circ + \cos^2 23^\circ \\ -\cos^2 23^\circ - \sin^2 23^\circ \\ -1 + \sin^2 23^\circ \end{pmatrix} \]

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\[ C_1 = \begin{pmatrix} 1 \\ -1 \\ \sin^2 23^\circ - 1 \end{pmatrix} \]

Thus, the determinant becomes:

\[ \Delta = \begin{vmatrix} 1 & \cos^2 23^\circ & -1 \\ -1 & -\sin^2 23^\circ & 1 \\ \sin^2 23^\circ - 1 & \cos^2 23^\circ & \cos^2 23^\circ \end{vmatrix} \]

### Step 3: Further simplify the determinant
Now, we can see that the second row can be simplified. We can also perform the operation \(C_1 \to C_1 + C_3\):

\[ C_1 = \begin{pmatrix} 1 + (-1) \\ -1 + 1 \\ (\sin^2 23^\circ - 1) + \cos^2 23^\circ \end{pmatrix} \]

This results in:

\[ C_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 4: Evaluate the determinant
Since one entire column of the determinant is now zero, we can conclude that the value of the determinant is:

\[ \Delta = 0 \]

### Final Answer
The value of the determinant is \(0\).

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