Flind the product of two matrices
`A =[[cos^(2) theta , cos theta sin theta],[cos theta sin theta ,sin^(2)theta]] B= [[cos^(2) phi,cos phi sin phi],[cos phisin phi,sin^(2)phi]]`
Show that, AB is the zero matrix if `theta and phi` differ by an
odd multipl of `pi/2`.

Explanation

To find the product of the two matrices \( A \) and \( B \) and show that \( AB \) is the zero matrix if \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \), we can follow these steps:

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

### Step 2: Compute the product \( AB \)
To compute the product \( AB \), we use the formula for matrix multiplication:
\[ AB = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \]
where \( a_{ij} \) are elements of matrix \( A \) and \( b_{ij} \) are elements of matrix \( B \).

Calculating each element of \( AB \):
- First row, first column:
\[ a_{11}b_{11} + a_{12}b_{21} = \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi \]
- First row, second column:
\[ a_{11}b_{12} + a_{12}b_{22} = \cos^2 \theta \cos \phi \sin \phi + \cos \theta \sin \theta \sin^2 \phi \]
- Second row, first column:
\[ a_{21}b_{11} + a_{22}b_{21} = \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi \]
- Second row, second column:
\[ a_{21}b_{12} + a_{22}b_{22} = \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi \]

Thus, we have:
\[ AB = \begin{bmatrix} \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi & \cos^2 \theta \cos \phi \sin \phi + \cos \theta \sin \theta \sin^2 \phi \\ \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi & \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi \end{bmatrix} \]

### Step 3: Simplify the expressions
Notice that:
1. The first element can be factored:
\[ \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi = \cos \phi \left( \cos^2 \theta \cos \phi + \sin \theta \sin \phi \right) \]
2. The second element can also be factored similarly.

### Step 4: Show that \( AB \) is the zero matrix
To show that \( AB \) is the zero matrix when \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \), we need to analyze the condition:
\[ \theta - \phi = (2n + 1) \frac{\pi}{2} \quad (n \in \mathbb{Z}) \]
This implies:
\[ \cos(\theta - \phi) = 0 \]
Thus, we have:
\[ \cos^2 \theta + \sin^2 \theta = 1 \quad \text{and} \quad \cos^2 \phi + \sin^2 \phi = 1 \]
This leads to:
\[ \cos \theta \cos \phi + \sin \theta \sin \phi = 0 \]
which implies that both rows and columns in \( AB \) will yield zero, confirming that \( AB \) is indeed the zero matrix.

### Conclusion
Thus, we have shown that \( AB \) is the zero matrix if \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \).