if `theta in R` then maximum value of `Delta =|{:(1,,1,,1),(1 ,,1+sintheta,,1),(1 ,, 1,,1+cos theta):}|` is

Options

A. `sqrt(3)//2)`
B. `1//2`
C. `1//sqrt(2)`
D. None of these

Explanation

To find the maximum value of the determinant \( \Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta \end{array} \right| \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ \Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta \end{array} \right| \]

### Step 2: Apply Column Operations
We can simplify the determinant by performing column operations. We will replace the second column \( C_2 \) with \( C_2 - C_1 \) and the third column \( C_3 \) with \( C_3 - C_1 \):
\[ \Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \sin \theta & 0 \\ 1 & 0 & \cos \theta \end{array} \right| \]

### Step 3: Expand the Determinant
Now we can expand the determinant along the first row:
\[ \Delta = 1 \cdot \left| \begin{array}{cc} \sin \theta & 0 \\ 0 & \cos \theta \end{array} \right| - 1 \cdot \left| \begin{array}{cc} 1 & 0 \\ 1 & \cos \theta \end{array} \right| + 1 \cdot \left| \begin{array}{cc} 1 & \sin \theta \\ 1 & 0 \end{array} \right| \]

Calculating the 2x2 determinants:
1. \( \left| \begin{array}{cc} \sin \theta & 0 \\ 0 & \cos \theta \end{array} \right| = \sin \theta \cos \theta \)
2. \( \left| \begin{array}{cc} 1 & 0 \\ 1 & \cos \theta \end{array} \right| = 1 \cdot \cos \theta - 0 \cdot 1 = \cos \theta \)
3. \( \left| \begin{array}{cc} 1 & \sin \theta \\ 1 & 0 \end{array} \right| = 1 \cdot 0 - 1 \cdot \sin \theta = -\sin \theta \)

Thus, we have:
\[ \Delta = \sin \theta \cos \theta - \cos \theta - \sin \theta \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:
\[ \Delta = \sin \theta \cos \theta - \cos \theta - \sin \theta \]
Factoring out \(-1\):
\[ \Delta = -(\cos \theta + \sin \theta - \sin \theta \cos \theta) \]

### Step 5: Find Maximum Value
To find the maximum value of \( \Delta \), we can rewrite \( \sin \theta \cos \theta \) using the double angle identity:
\[ \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \]
Thus,
\[ \Delta = -\left(\cos \theta + \sin \theta - \frac{1}{2} \sin 2\theta\right) \]

The maximum value of \( \sin 2\theta \) is 1, which occurs when \( 2\theta = \frac{\pi}{2} + 2k\pi \) for \( k \in \mathbb{Z} \). Therefore, the maximum value of \( \Delta \) occurs when:
\[ \Delta_{\text{max}} = -\left(1 + 0 - \frac{1}{2}\right) = -\frac{1}{2} \]

### Conclusion
The maximum value of \( \Delta \) is:
\[ \frac{1}{2} \]