If `A=(1)/(pi)[{:(sin^(-1)(pix),tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),cot^(-1)(pix)):}] and B=(1)/(pi) [{:(-cos^(-1)(pix), tan^(-1)((x)/(pi))),(sin^(-1)((x)/(pi)),-tan^(-1)(pix)):}]` then A-B is equal to

Options

A. I
B. `(1)/(2)I`
C. O
D. 2I

Answer

To solve the problem of finding \( A - B \) for the given matrices \( A \) and \( B \), we will follow these steps:

### Step 1: Write down the matrices \( A \) and \( B \)

Given:
\[ A = \frac{1}{\pi} \begin{pmatrix} \sin^{-1}(pix) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & \cot^{-1}(pix) \end{pmatrix} \]

\[ B = \frac{1}{\pi} \begin{pmatrix} -\cos^{-1}(pix) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & -\tan^{-1}(pix) \end{pmatrix} \]

### Step 2: Subtract the matrices \( A \) and \( B \)

Now, we will compute \( A - B \):
\[ A - B = \frac{1}{\pi} \begin{pmatrix} \sin^{-1}(pix) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & \cot^{-1}(pix) \end{pmatrix} - \frac{1}{\pi} \begin{pmatrix} -\cos^{-1}(pix) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ \sin^{-1}\left(\frac{x}{\pi}\right) & -\tan^{-1}(pix) \end{pmatrix} \]

### Step 3: Combine the matrices

Factoring out \( \frac{1}{\pi} \):
\[ A - B = \frac{1}{\pi} \left( \begin{pmatrix} \sin^{-1}(pix) + \cos^{-1}(pix) & 0 \\ 0 & \cot^{-1}(pix) + \tan^{-1}(pix) \end{pmatrix} \right) \]

### Step 4: Simplify the expressions

Using the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \):
\[ A - B = \frac{1}{\pi} \begin{pmatrix} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{pmatrix} \]

### Step 5: Factor out \( \frac{1}{\pi} \)

Now, we can factor out \( \frac{1}{\pi} \):
\[ A - B = \frac{1}{\pi} \cdot \frac{\pi}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 6: Final result

Thus, we have:
\[ A - B = \frac{1}{2} I \]
where \( I \) is the identity matrix.