If AB=0, then for the matrices
A=`[{:(cos^2theta,sinthetacostheta),(sinthetacostheta,cos^2theta):}]` and
B=`[{:(cos^2phi,sinphicosphi),(sinphicosphi,cos^2phi):}]`
`(theta-phi)`is

Options

A. an odd multiple of `(pi)/(2)`
B. an odd multiple of `pi`
C. an even multiple of `(pi)/(2)`
D. 0

Answer

To solve the problem, we need to find the value of \( \theta - \phi \) given that the product of matrices \( A \) and \( B \) is equal to the zero matrix. Let's break down the solution step by step.

### Step 1: Define the matrices
We have two matrices:
\[ A = \begin{pmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & \cos^2 \theta \end{pmatrix} \]
\[ B = \begin{pmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \cos^2 \phi \end{pmatrix} \]

### Step 2: Matrix multiplication
To find \( AB \), we multiply the matrices \( A \) and \( B \):
\[ AB = \begin{pmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & \cos^2 \theta \end{pmatrix} \begin{pmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \cos^2 \phi \end{pmatrix} \]

### Step 3: Calculate the elements of the resulting matrix
1. **First row, first column:**
\[ \cos^2 \theta \cdot \cos^2 \phi + \sin \theta \cos \theta \cdot \sin \phi \cos \phi \]
2. **First row, second column:**
\[ \cos^2 \theta \cdot \sin \phi \cos \phi + \sin \theta \cos \theta \cdot \cos^2 \phi \]
3. **Second row, first column:**
\[ \sin \theta \cos \theta \cdot \cos^2 \phi + \cos^2 \theta \cdot \sin \phi \cos \phi \]
4. **Second row, second column:**
\[ \sin \theta \cos \theta \cdot \sin \phi \cos \phi + \cos^2 \theta \cdot \cos^2 \phi \]

### Step 4: Set the resulting matrix equal to the zero matrix
Since \( AB = 0 \), we equate each element to zero:
1. \( \cos^2 \theta \cos^2 \phi + \sin \theta \cos \theta \sin \phi \cos \phi = 0 \)
2. \( \cos^2 \theta \sin \phi \cos \phi + \sin \theta \cos \theta \cos^2 \phi = 0 \)
3. \( \sin \theta \cos \theta \cos^2 \phi + \cos^2 \theta \sin \phi \cos \phi = 0 \)
4. \( \sin \theta \cos \theta \sin \phi \cos \phi + \cos^2 \theta \cos^2 \phi = 0 \)

### Step 5: Simplifying the equations
From the first equation, we can factor out common terms:
\[ \cos^2 \theta \cos^2 \phi + \sin \theta \cos \theta \sin \phi \cos \phi = 0 \]
This can be rearranged to:
\[ \cos^2 \theta \cos^2 \phi = -\sin \theta \cos \theta \sin \phi \cos \phi \]

### Step 6: Use trigonometric identities
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we can express \( \sin \theta \) in terms of \( \cos \theta \) and vice versa.

### Step 7: Solve for \( \theta - \phi \)
From the equations, we can derive:
\[ \cos(\theta - \phi) = 0 \]
This implies:
\[ \theta - \phi = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]
Thus, the solution can be expressed as:
\[ \theta - \phi = (2n + 1) \frac{\pi}{2}, \quad n \in \mathbb{Z} \]

### Final Answer:
\[ \theta - \phi = (2n + 1) \frac{\pi}{2} \]