The maximum value of `|(1,1,1),(1,1+sintheta,1),(1,1,1+costheta)|` is `1/2` Is it true or false

Explanation

To determine whether the maximum value of the determinant \( |(1,1,1),(1,1+\sin\theta,1),(1,1,1+\cos\theta)| \) is \( \frac{1}{2} \), we can follow these steps:

### Step 1: Write down the determinant
We start with the determinant:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin\theta & 1 \\ 1 & 1 & 1 + \cos\theta \end{vmatrix} \]

### Step 2: Apply row transformations
To simplify the determinant, we can perform row operations. We will transform \( R_1 \) to \( R_1 - R_2 \) and \( R_2 \) to \( R_2 - R_3 \):

\[ R_1 \rightarrow R_1 - R_2 \implies (1 - 1, 1 - (1 + \sin\theta), 1 - 1) = (0, -\sin\theta, 0) \]
\[ R_2 \rightarrow R_2 - R_3 \implies (1 - 1, (1 + \sin\theta) - 1, 1 - (1 + \cos\theta)) = (0, \sin\theta, -\cos\theta) \]

Now the determinant becomes:

\[ D = \begin{vmatrix} 0 & -\sin\theta & 0 \\ 0 & \sin\theta & -\cos\theta \\ 1 & 1 & 1 + \cos\theta \end{vmatrix} \]

### Step 3: Expand the determinant
We can expand the determinant along the first row:

\[ D = 0 \cdot \begin{vmatrix} \sin\theta & -\cos\theta \\ 1 & 1 + \cos\theta \end{vmatrix} - (-\sin\theta) \cdot \begin{vmatrix} 0 & -\cos\theta \\ 1 & 1 + \cos\theta \end{vmatrix} \]

The first term is zero, so we focus on the second term:

\[ D = \sin\theta \cdot \begin{vmatrix} 0 & -\cos\theta \\ 1 & 1 + \cos\theta \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} 0 & -\cos\theta \\ 1 & 1 + \cos\theta \end{vmatrix} = 0 \cdot (1 + \cos\theta) - (-\cos\theta) \cdot 1 = \cos\theta \]

Thus, we have:

\[ D = \sin\theta \cdot \cos\theta \]

### Step 4: Use the double angle identity
Using the double angle identity, we can express this as:

\[ D = \frac{1}{2} \sin(2\theta) \]

### Step 5: Find the maximum value
The maximum value of \( \sin(2\theta) \) is \( 1 \), which occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \). Therefore, the maximum value of \( D \) is:

\[ D_{\text{max}} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Conclusion
Thus, the statement that the maximum value of the determinant is \( \frac{1}{2} \) is **true**.

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