If `x+y+z=pi` then the value of `|{:(sin(x+y+z),sin(x+z),cosy),(-siny,0,tanx),(cos(x+z),tan(y+z),0):}|` is

Options

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

Answer

To solve the determinant given the condition \( x + y + z = \pi \), we will follow these steps:

### Step 1: Substitute \( x + y + z \) with \( \pi \)

Given the equation \( x + y + z = \pi \), we can substitute this into the determinant. The determinant we need to evaluate is:

\[ D = \begin{vmatrix} \sin(x+y+z) & \sin(x+z) & \cos y \\ -\sin y & 0 & \tan x \\ \cos(x+z) & \tan(y+z) & 0 \end{vmatrix} \]

### Step 2: Evaluate \( \sin(x+y+z) \)

Since \( x + y + z = \pi \):

\[ \sin(x+y+z) = \sin(\pi) = 0 \]

### Step 3: Evaluate \( \sin(x+z) \) and \( \cos(x+z) \)

We can express \( \sin(x+z) \) and \( \cos(x+z) \) using the identity:

\[ \sin(x+z) = \sin(\pi - y) = \sin y \]
\[ \cos(x+z) = \cos(\pi - y) = -\cos y \]

### Step 4: Evaluate \( \tan(y+z) \)

Similarly, we can express \( \tan(y+z) \):

\[ \tan(y+z) = \tan(\pi - x) = -\tan x \]

### Step 5: Substitute these values into the determinant

Now we substitute these values into the determinant:

\[ D = \begin{vmatrix} 0 & \sin y & \cos y \\ -\sin y & 0 & \tan x \\ -\cos y & -\tan x & 0 \end{vmatrix} \]

### Step 6: Expand the determinant

Using the determinant expansion formula, we can expand \( D \):

\[ D = 0 \cdot \begin{vmatrix} 0 & \tan x \\ -\tan x & 0 \end{vmatrix} - \sin y \cdot \begin{vmatrix} -\sin y & \tan x \\ -\cos y & 0 \end{vmatrix} + \cos y \cdot \begin{vmatrix} -\sin y & 0 \\ -\cos y & -\tan x \end{vmatrix} \]

### Step 7: Calculate the 2x2 determinants

Calculating the first 2x2 determinant:

\[ \begin{vmatrix} -\sin y & \tan x \\ -\cos y & 0 \end{vmatrix} = (-\sin y)(0) - (-\cos y)(\tan x) = \cos y \tan x \]

Calculating the second 2x2 determinant:

\[ \begin{vmatrix} -\sin y & 0 \\ -\cos y & -\tan x \end{vmatrix} = (-\sin y)(-\tan x) - (0)(-\cos y) = \sin y \tan x \]

### Step 8: Substitute back into the determinant

Now substituting back into \( D \):

\[ D = 0 - \sin y (\cos y \tan x) + \cos y (\sin y \tan x) \]

### Step 9: Simplify

This simplifies to:

\[ D = -\sin y \cos y \tan x + \cos y \sin y \tan x = 0 \]

### Final Answer

Thus, the value of the determinant is:

\[ \boxed{0} \]