The determinant `|(cos C,tan A,0),(sin B,0,-tan A),(0,sin B,cos C)|`
has the value, where A, B, C are angled of a triangle

Options

A. 0
B. 1
C. `sin A sin B`
D. `cos A cos B cos C`

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} \cos C & \tan A & 0 \\ \sin B & 0 & -\tan A \\ 0 & \sin B & \cos C \end{vmatrix} \]

we can use the property of determinants that allows us to expand along a row or column. In this case, we will expand along the first row.

### Step 1: Expand the determinant
Using the first row for expansion, we have:

\[ D = \cos C \cdot \begin{vmatrix} 0 & -\tan A \\ \sin B & \cos C \end{vmatrix} - \tan A \cdot \begin{vmatrix} \sin B & -\tan A \\ 0 & \cos C \end{vmatrix} + 0 \cdot \begin{vmatrix} \sin B & 0 \\ 0 & \sin B \end{vmatrix} \]

The last term is zero, so we can ignore it.

### Step 2: Calculate the first 2x2 determinant
Now we calculate the first 2x2 determinant:

\[ \begin{vmatrix} 0 & -\tan A \\ \sin B & \cos C \end{vmatrix} = (0 \cdot \cos C) - (-\tan A \cdot \sin B) = \tan A \sin B \]

### Step 3: Calculate the second 2x2 determinant
Next, we calculate the second 2x2 determinant:

\[ \begin{vmatrix} \sin B & -\tan A \\ 0 & \cos C \end{vmatrix} = (\sin B \cdot \cos C) - (0 \cdot -\tan A) = \sin B \cos C \]

### Step 4: Substitute back into the determinant
Substituting these results back into our expression for \(D\):

\[ D = \cos C \cdot (\tan A \sin B) - \tan A \cdot (\sin B \cos C) \]

### Step 5: Simplify the expression
Now, we simplify:

\[ D = \cos C \tan A \sin B - \tan A \sin B \cos C \]

Notice that both terms are identical but with opposite signs, thus:

\[ D = 0 \]

### Final Result
Therefore, the value of the determinant is:

\[ \boxed{0} \]