If `A + B + C =pi` , then the value of `|(sin(A+B+C),sinB,cosC),(-sinB,0,tanA),(cos(A+B),-tanA,0)|` is equal to

Options

A. 0
B. 1
C. `2 sin B tan A cos C`
D. none of these

Answer

To solve the given determinant problem, we start with the determinant:

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

Given that \( A + B + C = \pi \), we can substitute \( \sin(A+B+C) \) and \( \cos(A+B) \).

### Step 1: Substitute Values
Since \( A + B + C = \pi \), we have:

\[ \sin(A+B+C) = \sin(\pi) = 0 \]
\[ \cos(A+B) = \cos(\pi - C) = -\cos C \]

Now substituting these values into the determinant:

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

### Step 2: Calculate the Determinant
We can expand the determinant along the first row:

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

The first term is zero because it is multiplied by zero. Now we calculate the remaining two 2x2 determinants.

### Step 3: Calculate the 2x2 Determinants
1. For the first 2x2 determinant:

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

2. For the second 2x2 determinant:

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

### Step 4: Substitute Back
Now substituting these back into our expression for \( D \):

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

This simplifies to:

\[ D = -\sin B \cos C \tan A + \sin B \cos C \tan A = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{0} \]