If the max and values of
`Delta=|(1+sin^2x,cos^2x,sin2x),(sin^2x,1+cos^2x,sin2x),(sin^2x,cos^2x,1+sinx)|` and `alpha and beta` , then

Options

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

Answer

To solve the problem, we need to find the maximum and minimum values of the determinant:

\[ \Delta = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin x \end{vmatrix} \]

### Step 1: Simplify the Determinant
We can simplify the determinant by performing column operations. Let's add the first column and the second column together.

\[ C_1 \to C_1 + C_2 \]

This gives us:

\[ \Delta = \begin{vmatrix} 1 + \sin^2 x + \cos^2 x & \cos^2 x & \sin 2x \\ \sin^2 x + 1 + \cos^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x + \cos^2 x & \cos^2 x & 1 + \sin x \end{vmatrix} \]

Since \(\sin^2 x + \cos^2 x = 1\), we can simplify this further:

\[ \Delta = \begin{vmatrix} 2 & \cos^2 x & \sin 2x \\ 2 & 1 + \cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1 + \sin x \end{vmatrix} \]

### Step 2: Further Simplification
Next, we can perform row operations. Let's subtract the first row from the second row and the first row from the third row:

\[ R_2 \to R_2 - R_1 \]
\[ R_3 \to R_3 - \frac{1}{2} R_1 \]

This results in:

\[ \Delta = \begin{vmatrix} 2 & \cos^2 x & \sin 2x \\ 0 & 1 + \cos^2 x - \cos^2 x & \sin 2x - \sin 2x \\ 0 & \cos^2 x - \frac{1}{2} \cos^2 x & 1 + \sin x - 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 2 & \cos^2 x & \sin 2x \\ 0 & 1 & 0 \\ 0 & \frac{1}{2} \cos^2 x & \sin x \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now, we can calculate the determinant:

\[ \Delta = 2 \cdot \begin{vmatrix} 1 & 0 \\ \frac{1}{2} \cos^2 x & \sin x \end{vmatrix} = 2 \cdot (1 \cdot \sin x - 0 \cdot \frac{1}{2} \cos^2 x) = 2 \sin x \]

### Step 4: Find Maximum and Minimum Values
The maximum value of \(\sin x\) is \(1\) and the minimum value is \(-1\). Therefore:

- Maximum value (\(\alpha\)):
\[ \alpha = 2 \cdot 1 = 2 \]

- Minimum value (\(\beta\)):
\[ \beta = 2 \cdot (-1) = -2 \]

### Conclusion
Thus, the maximum and minimum values are:

\[ \alpha = 2, \quad \beta = -2 \]