If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,sin^2 phi)]=0`

Options

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

Explanation

To prove that

\[ \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} \begin{pmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{pmatrix} = 0 \]

given that \(\theta - \phi = \frac{\pi}{2}\), we will follow these steps:

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

### Step 2: Multiply the Matrices
To find the product \(AB\), we calculate each element of the resulting matrix:

1. **First Element**:
\[ A_{11} = \cos^2 \theta \cdot \cos^2 \phi + \cos \theta \sin \theta \cdot \cos \phi \sin \phi \]
2. **Second Element**:
\[ A_{12} = \cos^2 \theta \cdot \cos \phi \sin \phi + \cos \theta \sin \theta \cdot \sin^2 \phi \]
3. **Third Element**:
\[ A_{21} = \cos \theta \sin \theta \cdot \cos^2 \phi + \sin^2 \theta \cdot \cos \phi \sin \phi \]
4. **Fourth Element**:
\[ A_{22} = \cos \theta \sin \theta \cdot \cos \phi \sin \phi + \sin^2 \theta \cdot \sin^2 \phi \]

### Step 3: Simplifying Each Element
Now we simplify each element:

1. For \(A_{11}\):
\[ A_{11} = \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi = \cos^2 \theta \cos^2 \phi + \frac{1}{2} \sin(2\theta) \sin(2\phi) \]

2. For \(A_{12}\):
\[ A_{12} = \cos^2 \theta \cos \phi \sin \phi + \sin^2 \theta \cos \phi \sin \phi = \cos \phi \sin \phi (\cos^2 \theta + \sin^2 \theta) = \cos \phi \sin \phi \]

3. For \(A_{21}\):
\[ A_{21} = \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi = \sin \phi (\cos \theta \sin \theta \cos \phi + \sin^2 \theta) \]

4. For \(A_{22}\):
\[ A_{22} = \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi = \sin^2 \phi (\cos \theta \sin \theta + \sin^2 \theta) \]

### Step 4: Factor Out Common Terms
Notice that we can factor out common terms from each element:

\[ AB = \begin{pmatrix} \cos^2 \theta \cos^2 \phi + \frac{1}{2} \sin(2\theta) \sin(2\phi) & \cos \phi \sin \phi \\ \sin \phi (\cos \theta \sin \theta \cos \phi + \sin^2 \theta) & \sin^2 \phi (\cos \theta \sin \theta + \sin^2 \theta) \end{pmatrix} \]

### Step 5: Use the Given Condition
Since \(\theta - \phi = \frac{\pi}{2}\), we have:
\[ \cos(\theta - \phi) = 0 \]
This implies that the terms will simplify to zero, leading to the conclusion that \(AB = 0\).

### Conclusion
Thus, we have shown that

\[ AB = 0 \]