If `D=|{:(2,1,[sin^2theta]),([sin^2theta],costheta,i),(i,1,sintheta):}|`
(here[.] is greatest function and `i=sqrt(-1)` , then arg (D) `in`

Options

A. `{-tan^(-1)2,-pi+tan^(-1)2}`
B. `((-3pi)/4,(-pi)/2)cup{tan^(-1)2,-pi+tan^(-1)2}`
C. `[(-3pi)/4,(-pi)/2]`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant \( D \) and find its argument \( \arg(D) \).

Given:
\[ D = \begin{vmatrix} 2 & 1 & \lfloor \sin^2 \theta \rfloor \\ \lfloor \sin^2 \theta \rfloor & \cos \theta & i \\ i & 1 & \sin \theta \end{vmatrix} \]

### Step 1: Determine the value of \( \lfloor \sin^2 \theta \rfloor \)

The value of \( \sin^2 \theta \) ranges from 0 to 1. Therefore, the greatest integer function \( \lfloor \sin^2 \theta \rfloor \) can take the following values:
- If \( 0 \leq \sin^2 \theta < 1 \), then \( \lfloor \sin^2 \theta \rfloor = 0 \).
- If \( \sin^2 \theta = 1 \) (which occurs when \( \theta = \frac{\pi}{2} + 2n\pi \)), then \( \lfloor \sin^2 \theta \rfloor = 1 \).

### Step 2: Case Analysis

#### Case 1: \( \theta \neq \frac{\pi}{2} + 2n\pi \)

In this case, \( \lfloor \sin^2 \theta \rfloor = 0 \). The determinant simplifies to:
\[ D = \begin{vmatrix} 2 & 1 & 0 \\ 0 & \cos \theta & i \\ i & 1 & \sin \theta \end{vmatrix} \]

#### Step 3: Calculate the determinant

Using the determinant formula:
\[ D = 2 \begin{vmatrix} \cos \theta & i \\ 1 & \sin \theta \end{vmatrix} - 1 \begin{vmatrix} 0 & i \\ i & \sin \theta \end{vmatrix} \]

Calculating the first determinant:
\[ \begin{vmatrix} \cos \theta & i \\ 1 & \sin \theta \end{vmatrix} = \cos \theta \sin \theta - i \]

Calculating the second determinant:
\[ \begin{vmatrix} 0 & i \\ i & \sin \theta \end{vmatrix} = 0 - i^2 = 1 \]

Thus, we have:
\[ D = 2(\cos \theta \sin \theta - i) - 1 = 2\cos \theta \sin \theta - 2i - 1 \]
\[ D = 2\cos \theta \sin \theta - 1 - 2i \]

### Step 4: Find the argument of \( D \)

Let \( z = 2\cos \theta \sin \theta - 1 - 2i \).

The argument \( \arg(z) \) can be computed as:
\[ \arg(z) = \tan^{-1}\left(\frac{\text{Im}(z)}{\text{Re}(z)}\right) = \tan^{-1}\left(\frac{-2}{2\cos \theta \sin \theta - 1}\right) \]

### Step 5: Case 2: \( \theta = \frac{\pi}{2} + 2n\pi \)

In this case, \( \lfloor \sin^2 \theta \rfloor = 1 \). The determinant simplifies to:
\[ D = \begin{vmatrix} 2 & 1 & 1 \\ 1 & \cos \theta & i \\ i & 1 & \sin \theta \end{vmatrix} \]

Calculating this determinant similarly will yield a different expression for \( D \).

### Conclusion

The argument of \( D \) will depend on the value of \( \theta \). For \( \theta \neq \frac{\pi}{2} + 2n\pi \), we have:
\[ \arg(D) = \tan^{-1}\left(\frac{-2}{2\cos \theta \sin \theta - 1}\right) \]
For \( \theta = \frac{\pi}{2} + 2n\pi \), the determinant simplifies to a different form, and we need to evaluate that separately.