If `x epsilon R` the determinant
`Delta=|(1,cosx,0),(-1,1-cosx,sinx+cosx),(0,-1,1-sqrt(2)sin(x+pi//4))|` equals

Options

A. 0
B. `-1`
C. 1
D. None of these

Answer

To solve the determinant

\[ \Delta = \begin{vmatrix} 1 & \cos x & 0 \\ -1 & 1 - \cos x & \sin x + \cos x \\ 0 & -1 & 1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \end{vmatrix} \]

we will use the method of cofactor expansion along the first row.

### Step 1: Expand the determinant using the first row

The determinant can be expanded as follows:

\[ \Delta = 1 \cdot \begin{vmatrix} 1 - \cos x & \sin x + \cos x \\ -1 & 1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \end{vmatrix} - \cos x \cdot \begin{vmatrix} -1 & \sin x + \cos x \\ 0 & 1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \end{vmatrix} + 0 \]

### Step 2: Calculate the first 2x2 determinant

Calculating the first determinant:

\[ \begin{vmatrix} 1 - \cos x & \sin x + \cos x \\ -1 & 1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \end{vmatrix} = (1 - \cos x)(1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) - (-1)(\sin x + \cos x) \]

This simplifies to:

\[ (1 - \cos x)(1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) + \sin x + \cos x \]

### Step 3: Calculate the second 2x2 determinant

Calculating the second determinant:

\[ \begin{vmatrix} -1 & \sin x + \cos x \\ 0 & 1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \end{vmatrix} = -1 \cdot (1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) - 0 \]

This simplifies to:

\[ - (1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) \]

### Step 4: Substitute back into the determinant expression

Now substituting back into the expression for \(\Delta\):

\[ \Delta = 1 \cdot \left[(1 - \cos x)(1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) + \sin x + \cos x\right] + \cos x \cdot (1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) \]

### Step 5: Combine terms

Combining the terms gives:

\[ \Delta = (1 - \cos x)(1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) + \sin x + \cos x + \cos x(1 - \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) \]

### Step 6: Simplify the expression

Now, we can simplify this expression further, but we can also observe that certain terms may cancel out or combine nicely.

After careful observation and simplification, we find that:

\[ \Delta = 1 \]

### Final Answer

Thus, the value of the determinant \(\Delta\) is:

\[ \Delta = 1 \]