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 `I` (b) 0 (c) `2I` (d) `1/2I`

Explanation

To solve the problem, we need to compute the difference between the matrices \( A \) and \( B \) and determine what that difference equals.

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

\[
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