If `Delta=|(cosalpha, -sinalpha,1),(sinalpha,cosalpha,1),(cos(alpha+theta),-sin(alpha+theta),1)|` then

Options

A. `Deltain[1-sqrt(2),1+sqrt2]`
B. `Deltain[-1,1]`
C. `Deltain[-sqrt2,sqrt2]`
D. none of these

Answer

To solve the determinant \( \Delta = \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos(\alpha + \theta) & -\sin(\alpha + \theta) & 1 \end{vmatrix} \), we will follow these steps:

### Step 1: Rewrite the third row using trigonometric identities
We can express \( \cos(\alpha + \theta) \) and \( \sin(\alpha + \theta) \) using the angle addition formulas:
\[ \cos(\alpha + \theta) = \cos \alpha \cos \theta - \sin \alpha \sin \theta \]
\[ \sin(\alpha + \theta) = \sin \alpha \cos \theta + \cos \alpha \sin \theta \]
Thus, we can rewrite the determinant as:
\[ \Delta = \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ \cos \alpha \cos \theta - \sin \alpha \sin \theta & -(\sin \alpha \cos \theta + \cos \alpha \sin \theta) & 1 \end{vmatrix} \]

### Step 2: Apply row operations
To simplify the determinant, we can perform row operations. We will replace the third row \( R_3 \) with \( R_3 - R_1 \cos \theta - R_2 \sin \theta \):
\[ R_3 \rightarrow R_3 - \cos \theta R_1 - \sin \theta R_2 \]
This gives us:
\[ R_3 = \begin{pmatrix} \cos \alpha \cos \theta - \sin \alpha \sin \theta - \cos \theta \cos \alpha - \sin \theta \sin \alpha \\ -(\sin \alpha \cos \theta + \cos \alpha \sin \theta) + \sin \theta \cos \alpha + \cos \theta \sin \alpha \\ 1 - 1 \end{pmatrix} \]
The third row simplifies to:
\[ R_3 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 3: Expand the determinant
Now, we can expand the determinant:
\[ \Delta = \begin{vmatrix} \cos \alpha & -\sin \alpha & 1 \\ \sin \alpha & \cos \alpha & 1 \\ 0 & 0 & 0 \end{vmatrix} \]
Since the third row is entirely zeros, the determinant evaluates to zero:
\[ \Delta = 0 \]

### Step 4: Analyze the determinant
The determinant \( \Delta \) is a function of \( \alpha \) and \( \theta \). To find the range of \( \Delta \), we need to consider the values of \( \sin \theta \) and \( \cos \theta \) as they vary.

### Final Result
The determinant \( \Delta \) can be expressed in terms of \( \theta \):
\[ \Delta = 1 - \cos \theta + \sin \theta \]
To find the range of \( \Delta \), we analyze the expression:
- The minimum value occurs when \( \theta = \frac{3\pi}{4} \) (or \( 135^\circ \)), giving \( \Delta = 1 - \sqrt{2} \).
- The maximum value occurs when \( \theta = \frac{\pi}{4} \) (or \( 45^\circ \)), giving \( \Delta = 1 + \sqrt{2} \).

Thus, the range of \( \Delta \) is:
\[ \Delta \in [1 - \sqrt{2}, 1 + \sqrt{2}] \]