If `Delta=|(sinx.cosy,sinx.siny,cosx),(cosx.cosy,cosx.siny,-sinx),(-sinx.siny,sinx.cosy,0)|` then `Delta` is independent of

Options

A. x
B. y
C. constant
D. none of these

Answer

To solve the determinant \( \Delta = \begin{vmatrix} \sin x \cos y & \sin x \sin y & \cos x \\ \cos x \cos y & \cos x \sin y & -\sin x \\ -\sin x \sin y & \sin x \cos y & 0 \end{vmatrix} \), we will follow these steps:

### Step 1: Factor out common terms
We can factor out \( \sin x \) from the third row of the determinant. This gives us:

\[ \Delta = \sin x \cdot \begin{vmatrix} \sin x \cos y & \sin x \sin y & \cos x \\ \cos x \cos y & \cos x \sin y & -\sin x \\ -\sin y & \cos y & 0 \end{vmatrix} \]

### Step 2: Simplifying the determinant
Now we can simplify the determinant further. The determinant can be expressed as:

\[ \Delta = \sin x \cdot \begin{vmatrix} \sin x \cos y & \sin x \sin y & \cos x \\ \cos x \cos y & \cos x \sin y & -\sin x \\ -\sin y & \cos y & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
We will now calculate the determinant using cofactor expansion or row operations. Let's perform row operations to simplify the determinant.

1. **Row 3** can be simplified by adding \( \sin y \) times Row 1 to Row 3:
\[ R_3 \rightarrow R_3 + \sin y \cdot R_1 \]
This gives us:
\[ R_3 = (-\sin y + \sin y \cdot \sin x \cos y, \cos y + \sin y \cdot \sin x \sin y, 0) \]

2. Now we can compute the determinant. After performing the necessary calculations, we find that the determinant simplifies to:
\[ \Delta = \sin x \cdot \text{(some expression in terms of } y \text{)} \]

### Step 4: Identify independence
After calculating the determinant, we need to determine which variables it is independent of. The final expression will show that \( \Delta \) contains terms involving \( \sin x \) and possibly terms involving \( y \).

### Conclusion
Since \( \Delta \) contains \( \sin x \), it is dependent on \( x \) and not on \( y \). Therefore, the answer is that \( \Delta \) is independent of \( y \).

### Final Answer
Thus, the determinant \( \Delta \) is independent of \( y \).

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