`" If "|{:(sin theta cos phi,,sin theta sin phi,,cos theta),(cos theta cos phi,, cos theta sin phi,,-sin theta),(-sin theta sin phi,,sin theta cos phi,,theta):}|`then

Options

A. `Delta` is independent of theta
B. `Delta` is independent of `phi`
C. `Delta `is a constant
D. `[(dDelta)/d (theta)]_(theta=pi//2)=0`

Explanation

To solve the given determinant problem, we will follow a step-by-step approach. The determinant is given as:

\[ \Delta = \begin{vmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & \theta \end{vmatrix} \]

### Step 1: Perform Column Operation
We will perform the column operation \( C_1 \leftarrow C_1 - \frac{\cos \phi}{\sin \phi} C_2 \).

\[ C_1 = \sin \theta \cos \phi - \frac{\cos \phi}{\sin \phi} (\sin \theta \sin \phi) \]
This simplifies to:
\[ C_1 = \sin \theta \cos \phi - \cos \phi \sin \theta = 0 \]

Thus, the first column becomes:
\[ \begin{pmatrix} 0 \\ 0 \\ -\sin \theta \sin \phi \end{pmatrix} \]

### Step 2: Update the Determinant
After the column operation, the determinant becomes:
\[ \Delta = \begin{vmatrix} 0 & \sin \theta \sin \phi & \cos \theta \\ 0 & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & \theta \end{vmatrix} \]

### Step 3: Calculate the Determinant
Since the first column is all zeros, we can expand the determinant along the first column:
\[ \Delta = 0 \]

### Step 4: Analyze the Result
Since the determinant evaluates to zero, it indicates that the determinant is independent of both \(\theta\) and \(\phi\). Therefore, we can conclude:

1. \(\Delta\) is independent of \(\theta\).
2. \(\Delta\) is independent of \(\phi\).
3. \(\Delta\) is a constant (specifically, zero).

### Step 5: Derivative with Respect to \(\theta\)
To find \( \frac{d\Delta}{d\theta} \):
\[ \frac{d\Delta}{d\theta} = 0 \]
At \(\theta = \frac{\pi}{2}\), this is also zero.

### Conclusion
The correct options are:
- \(\Delta\) is independent of \(\phi\)
- \(\frac{d\Delta}{d\theta} \text{ at } \theta = \frac{\pi}{2} = 0\)