If `|[cos(A+B), -sin(A+B), cos2B], [sinA, cosA, sinB], [-cosA, sinA, cosB]|=0`, then B=

Options

A. `npi, ninZ`
B. `2npi, ninZ`
C. `(2npm1)(pi)/(2), ninZ`
D. `(2npipm1)pi, ninZ`

Answer

To solve the given problem, we need to evaluate the determinant and set it equal to zero. Let's break down the steps:

### Step 1: Write the Determinant
We are given the determinant:

\[ D = \begin{vmatrix} \cos(A+B) & -\sin(A+B) & \cos(2B) \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \end{vmatrix} \]

### Step 2: Expand the Determinant
We will expand the determinant along the first row:

\[ D = \cos(A+B) \begin{vmatrix} \cos A & \sin B \\ \sin A & \cos B \end{vmatrix} - (-\sin(A+B)) \begin{vmatrix} \sin A & \sin B \\ -\cos A & \cos B \end{vmatrix} + \cos(2B) \begin{vmatrix} \sin A & \cos A \\ -\cos A & \sin A \end{vmatrix} \]

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

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

2. For the second determinant:

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

3. For the third determinant:

\[ \begin{vmatrix} \sin A & \cos A \\ -\cos A & \sin A \end{vmatrix} = \sin^2 A + \cos^2 A = 1 \]

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

\[ D = \cos(A+B) \cdot \cos(A+B) + \sin(A+B) \cdot \sin(A+B) + \cos(2B) \cdot 1 \]

This simplifies to:

\[ D = \cos^2(A+B) + \sin^2(A+B) + \cos(2B) \]

Using the Pythagorean identity \(\cos^2 x + \sin^2 x = 1\):

\[ D = 1 + \cos(2B) \]

### Step 5: Set the Determinant Equal to Zero
We set the determinant equal to zero:

\[ 1 + \cos(2B) = 0 \]

This simplifies to:

\[ \cos(2B) = -1 \]

### Step 6: Solve for B
The cosine function equals -1 at odd multiples of \(\pi\):

\[ 2B = (2n + 1)\pi \quad \text{where } n \in \mathbb{Z} \]

Dividing both sides by 2 gives:

\[ B = \frac{(2n + 1)\pi}{2} = n\pi + \frac{\pi}{2} \]

This means \(B\) can be expressed as:

\[ B = \frac{\pi}{2} + n\pi \quad \text{where } n \in \mathbb{Z} \]

### Final Answer
Thus, the value of \(B\) is:

\[ B = n\pi + \frac{\pi}{2} \quad \text{where } n \in \mathbb{Z} \]