If $A=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(\pi \mathrm{x}) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi \mathrm{x})\end{array}\right], B=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$, then $A-B$ is equal to
(a) $I$
(b) 0
(c) 21
(d) $\frac{1}{2} I$
Disclaimer: There is a misprint in the question. $\operatorname{Cos}^{-1}$ should be written instead of $\operatorname{Cot}^{-1}$.
Given: $A=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(\pi \mathrm{x}) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi \mathrm{x})\end{array}\right], B=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$
$A-B=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(\pi \mathrm{x}) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi \mathrm{x})\end{array}\right]-\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$
$=\frac{1}{\pi}\left(\left[\begin{array}{cc}\sin ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]+\left[\begin{array}{cc}\cos ^{-1}(\pi x) & -\tan ^{-1}\left(\frac{x}{\pi}\right) \\ -\sin ^{-1}\left(\frac{x}{\pi}\right) & \tan ^{-1}(\pi x)\end{array}\right]\right)$
$=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(\pi x)+\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right)-\tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right)-\sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x)\end{array}\right]$
$=\frac{1}{\pi}\left[\begin{array}{cc}\frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2}\end{array}\right]$ $\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right.$ and $\left.\cot ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}\right)$
$=\frac{1}{\pi} \times \frac{\pi}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\frac{1}{2} I$
Hence, the correct option is (d).
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