Find the maximum value of `|(1,1,1),(1,1+ sin theta, 1),(1,1, 1+ sin theta)|`.

Answer

To find the maximum value of the determinant \( D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \sin \theta \end{vmatrix} \), we can follow these steps:

### Step 1: Apply Column Operations
We can simplify the determinant by applying column operations. We will perform the following operations:
- Replace column \( C_1 \) with \( C_1 - C_2 \)
- Replace column \( C_2 \) with \( C_2 - C_3 \)

This gives us:
\[ D = \begin{vmatrix} 1 - (1 + \sin \theta) & 1 & 1 \\ 1 - 1 & 1 + \sin \theta - 1 & 1 \\ 1 - 1 & 1 & 1 + \sin \theta - 1 \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} -\sin \theta & 1 & 1 \\ 0 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \end{vmatrix} \]

### Step 2: Calculate the Determinant
Now we can calculate the determinant of the simplified matrix:
\[ D = -\sin \theta \begin{vmatrix} \sin \theta & 1 \\ 1 & \sin \theta \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} \sin \theta & 1 \\ 1 & \sin \theta \end{vmatrix} = \sin^2 \theta - 1 \]
Thus, we have:
\[ D = -\sin \theta (\sin^2 \theta - 1) = -\sin \theta (\sin^2 \theta - 1) = -\sin \theta (-\cos^2 \theta) = \sin \theta \cos^2 \theta \]

### Step 3: Find the Maximum Value
To find the maximum value of \( D = \sin \theta \cos^2 \theta \), we can use the fact that \( \sin \theta \) and \( \cos \theta \) are bounded between -1 and 1.

Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can express \( D \) in terms of \( \sin \theta \):
\[ D = \sin \theta (1 - \sin^2 \theta) \]
Let \( x = \sin \theta \), then:
\[ D = x(1 - x^2) = x - x^3 \]

### Step 4: Differentiate to Find Critical Points
To find the maximum value, we differentiate \( D \) with respect to \( x \):
\[ \frac{dD}{dx} = 1 - 3x^2 \]
Setting the derivative to zero to find critical points:
\[ 1 - 3x^2 = 0 \Rightarrow 3x^2 = 1 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \frac{1}{\sqrt{3}} \]

### Step 5: Evaluate the Maximum Value
Now we evaluate \( D \) at \( x = \frac{1}{\sqrt{3}} \):
\[ D = \frac{1}{\sqrt{3}} \left( 1 - \left(\frac{1}{\sqrt{3}}\right)^2 \right) = \frac{1}{\sqrt{3}} \left( 1 - \frac{1}{3} \right) = \frac{1}{\sqrt{3}} \cdot \frac{2}{3} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9} \]

### Step 6: Check Endpoints
We also need to check the endpoints \( x = 0 \) and \( x = 1 \):
- For \( x = 0 \), \( D = 0 \)
- For \( x = 1 \), \( D = 1 \cdot 0 = 0 \)

Thus, the maximum value of \( D \) occurs at \( x = \frac{1}{\sqrt{3}} \).

### Final Answer
The maximum value of \( |D| \) is \( \frac{2\sqrt{3}}{9} \).