What is the value of the following determinants?
`|{:(cosC, tanA, 0),(sinB,0,-tanA),(0,sinB,cosC):}|`

Options

A. `-1`
B. 0
C. 2tanAsinBsinC
D. `-2tanAsinBsinC`

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 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:

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

The third term is zero because it is multiplied by zero.

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

Calculating the first 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

Calculating the second 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 expression

Now substituting these values back into the determinant expression:

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

### Step 5: Simplify the expression

This simplifies to:

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

### Step 6: Final result

Both terms are equal but opposite in sign, so they cancel each other out:

\[ D = 0 \]

Thus, the value of the determinant is

\[ \boxed{0} \]