If the maximum and minimum values of the determinant
`|(1 + sin^(2)x,cos^(2) x,sin 2x),(sin^(2) x,1 + cos^(2) x,sin 2x),(sin^(2) x,cos^(2) x,1 + sin 2x)|` are `alpha and beta`, then

Options

A. `alpha + beta^(99) = 4`
B. `alpha^(3) - beta^(17) = 26`
C. `alpha^(2n) - beta^(2n)` is always even integer for `n in N`
D. a triangle can be constructed having its sides as `alpha, beta and alpha - beta`

Answer

To solve the given determinant problem step by step, we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant given in the question:
\[ D = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

### Step 2: Simplify the Determinant
To simplify the determinant, we can perform column operations. We will add the first two columns together:
\[ C_1 \rightarrow C_1 + C_2 \]
This gives us:
\[ D = \begin{vmatrix} (1 + \sin^2 x + \cos^2 x) & \cos^2 x & \sin 2x \\ (\sin^2 x + 1 + \cos^2 x) & 1 + \cos^2 x & \sin 2x \\ (\sin^2 x + \cos^2 x) & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]
Using the identity \(\sin^2 x + \cos^2 x = 1\), we can simplify further:
\[ D = \begin{vmatrix} 2 + \sin^2 x & \cos^2 x & \sin 2x \\ 2 & 1 + \cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant using the first row:
\[ D = (2 + \sin^2 x) \begin{vmatrix} 1 + \cos^2 x & \sin 2x \\ \cos^2 x & 1 + \sin 2x \end{vmatrix} - \cos^2 x \begin{vmatrix} 2 & \sin 2x \\ 1 & 1 + \sin 2x \end{vmatrix} + \sin 2x \begin{vmatrix} 2 & 1 + \cos^2 x \\ 1 & \cos^2 x \end{vmatrix} \]

### Step 4: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:
1. \(\begin{vmatrix} 1 + \cos^2 x & \sin 2x \\ \cos^2 x & 1 + \sin 2x \end{vmatrix} = (1 + \cos^2 x)(1 + \sin 2x) - \sin 2x \cos^2 x\)

2. \(\begin{vmatrix} 2 & \sin 2x \\ 1 & 1 + \sin 2x \end{vmatrix} = 2(1 + \sin 2x) - \sin 2x = 2 + 2\sin 2x - \sin 2x = 2 + \sin 2x\)

3. \(\begin{vmatrix} 2 & 1 + \cos^2 x \\ 1 & \cos^2 x \end{vmatrix} = 2\cos^2 x - (1 + \cos^2 x) = 2\cos^2 x - 1 - \cos^2 x = \cos^2 x - 1\)

### Step 5: Substitute Back into the Determinant
Substituting these back into the expression for \(D\):
\[ D = (2 + \sin^2 x) \left[(1 + \cos^2 x)(1 + \sin 2x) - \sin 2x \cos^2 x\right] - \cos^2 x (2 + \sin 2x) + \sin 2x (\cos^2 x - 1) \]

### Step 6: Find Maximum and Minimum Values
To find the maximum and minimum values of \(D\), we analyze the expression. The maximum value occurs when \(\sin^2 x\) and \(\cos^2 x\) take their maximum values, leading to:
\[ \alpha = 3 \quad \text{(maximum)} \]
And the minimum value occurs when these values are minimized, leading to:
\[ \beta = 1 \quad \text{(minimum)} \]

### Final Result
Thus, the maximum and minimum values of the determinant are:
\[ \alpha = 3, \quad \beta = 1 \]