If `A=1/pi[sin^(-1)(pix)tan^(-1)(x/pi)sin^(-1)(x/pi)cot^(-1)(pix)]` and `B=1/pi[-cot^(-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`

Options

A. I
B. 0
C. 2I
D. `(1)/(2)I`

Explanation

To solve the problem, we need to compute \( A - B \) where:

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

\[
B = \frac{1}{\pi} \begin{bmatrix} -\cot^{-1}(\pi x) & \tan^{-1}\left(\frac{x}{\pi}\right) \\ -\tan^{-1}(\pi x) & \sin^{-1}\left(\frac{x}{\pi}\right) \end{bmatrix}
...