If `A=[(cos^(2) theta, cos theta sin theta),(cos theta sin theta, sin^(2)theta)]`
and `B=[(cos^(2) phi, cos phi sin phi),(cos phi sin phi, sin^(2)phi)]` are the two matrices such that the product AB is the null matrix then `theta-phi` is equal to

Options

A. 0
B. multiple of `pi`
C. on odd multiple of `pi//2`
D. None of these

Answer

To solve the problem, we need to find the value of \( \theta - \phi \) given that the product of the matrices \( A \) and \( B \) is the null matrix.

Given:
\[ A = \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} \]
\[ B = \begin{pmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{pmatrix} \]

We need to compute the product \( AB \) and set it equal to the null matrix.

### Step 1: Calculate the product \( AB \)

The product of two matrices \( A \) and \( B \) is given by:
\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \]

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

### Step 2: Simplify the elements of \( AB \)

1. First element:
\[ \cos^2 \theta \cdot \cos^2 \phi + \cos \theta \sin \theta \cdot \cos \phi \sin \phi = \cos^2 \theta \cos^2 \phi + \frac{1}{2} \sin(2\theta) \sin(2\phi) \]

2. Second element:
\[ \cos^2 \theta \cdot \cos \phi \sin \phi + \cos \theta \sin \theta \cdot \sin^2 \phi = \frac{1}{2} \sin(2\theta) \sin(2\phi) + \cos^2 \theta \sin^2 \phi \]

3. Third element:
\[ \cos \theta \sin \theta \cdot \cos^2 \phi + \sin^2 \theta \cdot \cos \phi \sin \phi = \frac{1}{2} \sin(2\theta) \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi \]

4. Fourth element:
\[ \cos \theta \sin \theta \cdot \cos \phi \sin \phi + \sin^2 \theta \cdot \sin^2 \phi = \frac{1}{2} \sin(2\theta) \sin(2\phi) + \sin^2 \theta \sin^2 \phi \]

### Step 3: Set the product equal to the null matrix

Setting \( AB = 0 \):
\[ \begin{pmatrix} \cos^2 \theta \cos^2 \phi + \frac{1}{2} \sin(2\theta) \sin(2\phi) & \frac{1}{2} \sin(2\theta) \sin(2\phi) + \cos^2 \theta \sin^2 \phi \\ \frac{1}{2} \sin(2\theta) \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi & \frac{1}{2} \sin(2\theta) \sin(2\phi) + \sin^2 \theta \sin^2 \phi \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Step 4: Analyze the equations

From the first element:
\[ \cos^2 \theta \cos^2 \phi + \frac{1}{2} \sin(2\theta) \sin(2\phi) = 0 \]

This implies:
\[ \cos^2 \theta \cos^2 \phi = -\frac{1}{2} \sin(2\theta) \sin(2\phi) \]

### Step 5: Use trigonometric identities

Using the identity \( \cos(x - y) = \cos x \cos y + \sin x \sin y \), we can relate the equations to \( \cos(\theta - \phi) \).

### Step 6: Solve for \( \theta - \phi \)

Since \( AB = 0 \), we can conclude:
\[ \cos(\theta - \phi) = 0 \]
This means:
\[ \theta - \phi = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

Thus, \( \theta - \phi \) can be expressed as:
\[ \theta - \phi = (2n + 1) \frac{\pi}{2} \]

### Final Answer
The value of \( \theta - \phi \) is an odd multiple of \( \frac{\pi}{2} \).

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