If \( A=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{I I}\right) \\ \tan ^{-1}\left(\frac{x}{I I}\right) & \cot ^{-1}(\Pi x)\end{array}\right] \) then \( A-B \) is

Given that,
\( A=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(I X) & \tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{\Pi}\right) & \cot ^{-1}(\Pi X)\end{array}\right] \)
and
\( B=\left[\begin{array}{ll}-\cos ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{\Pi}\right) & -\tan ^{-1}(\Pi x)\end{array}\right] \)
So,
\( A-B=\frac{1}{\Pi}\left[\begin{array}{ll}\sin ^{-1}(\Pi x)+\cos ^{-1}(\Pi x) & \tan ^{-1}\left(\frac{x}{\Pi}\right)-\tan ^{-1}\left(\frac{x}{\Pi}\right) \\ \sin ^{-1}\left(\frac{x}{I I}\right)-\sin ^{-1}\left(\frac{x}{I I}\right) & \cot ^{-1}(\Pi x)+\tan ^{-1}(\Pi x)\end{array}\right] \)
Since, \( \sin ^{-1} x+\cos ^{-1} x=\frac{\Pi}{2} \) and \( \tan ^{-1} x+\cot ^{-1} x=\frac{I}{2} . \) So,
\( A-B=\frac{1}{\Pi}\left[\begin{array}{ccc}\frac{I I}{2} & 0 \\ 0 & \frac{\Pi}{2}\end{array}\right]=\frac{1}{\Pi} \times \frac{\Pi}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
\( =\frac{\Pi}{2} I \times \frac{1}{\Pi}=\frac{1}{2} I \)