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. I
B. 0
C. `-I`
D. None of these

Answer

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 can follow these steps:

### Step 1: Express \(\theta\) in terms of \(\phi\)
From the equation \(\theta - \phi = \frac{\pi}{2}\), we can express \(\theta\) as:
\[ \theta = \frac{\pi}{2} + \phi \]

### Step 2: Find \(\cos \theta\) and \(\sin \theta\)
Using the angle addition formulas:
\[ \cos \theta = \cos\left(\frac{\pi}{2} + \phi\right) = -\sin \phi \]
\[ \sin \theta = \sin\left(\frac{\pi}{2} + \phi\right) = \cos \phi \]

### Step 3: Substitute \(\cos \theta\) and \(\sin \theta\) into the first matrix
Substituting these values into the first matrix, we get:
\[ \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} = \begin{pmatrix} (-\sin \phi)^2 & (-\sin \phi)(\cos \phi) \\ (-\sin \phi)(\cos \phi) & (\cos \phi)^2 \end{pmatrix} = \begin{pmatrix} \sin^2 \phi & -\sin \phi \cos \phi \\ -\sin \phi \cos \phi & \cos^2 \phi \end{pmatrix} \]

### Step 4: Multiply the two matrices
Now we need to multiply the two matrices:
\[ \begin{pmatrix} \sin^2 \phi & -\sin \phi \cos \phi \\ -\sin \phi \cos \phi & \cos^2 \phi \end{pmatrix} \begin{pmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{pmatrix} \]

Calculating the elements of the resulting matrix:

- **Element (1,1)**:
\[ \sin^2 \phi \cdot \cos^2 \phi + (-\sin \phi \cos \phi) \cdot (\cos \phi \sin \phi) = \sin^2 \phi \cos^2 \phi - \sin^2 \phi \cos^2 \phi = 0 \]

- **Element (1,2)**:
\[ \sin^2 \phi \cdot (\cos \phi \sin \phi) + (-\sin \phi \cos \phi) \cdot \sin^2 \phi = \sin^2 \phi \cos \phi \sin \phi - \sin^2 \phi \cos \phi = 0 \]

- **Element (2,1)**:
\[ (-\sin \phi \cos \phi) \cdot \cos^2 \phi + \cos^2 \phi \cdot (-\sin \phi \cos \phi) = -\sin \phi \cos^3 \phi - \sin \phi \cos^3 \phi = -2\sin \phi \cos^3 \phi \]

- **Element (2,2)**:
\[ (-\sin \phi \cos \phi) \cdot (\cos \phi \sin \phi) + \cos^2 \phi \cdot \sin^2 \phi = -\sin^2 \phi \cos^2 \phi + \cos^2 \phi \sin^2 \phi = 0 \]

### Step 5: Combine the results
Combining these results, we find that all elements of the resulting matrix are zero:
\[ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0 \]

### Conclusion
Thus, we have shown that:
\[ \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} \begin{pmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{pmatrix} = 0 \]