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

Options

A. `(2n + 1)pi//2, n in Z`
B. `npi, n in Z`
C. `(2n + 1) pi, n in Z`
D. `2n pi, n inZ`

Explanation

To solve the problem, we need to evaluate the determinant of the given matrix and set it equal to zero. The matrix is:

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

### Step 1: Expand the Determinant

We will expand the determinant along the first row (R1):

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

### Step 2: Calculate the 2x2 Determinants

1. **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. **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. **Third Determinant:**
\[ \begin{vmatrix} \sin A & \cos A \\ -\cos A & \sin A \end{vmatrix} = \sin^2 A + \cos^2 A = 1 \]

### Step 3: Substitute Back into the Determinant Expression

Now substituting these back into the determinant:

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

This simplifies to:

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

Using the Pythagorean identity:

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

### Step 4: Set the Determinant Equal to Zero

We set the determinant equal to zero:

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

This implies:

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

### Step 5: Solve for B

The cosine function equals -1 at odd multiples of \(\pi\):

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

Thus, dividing by 2 gives:

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

### Final Answer

The solution for \(B\) is:

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