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If $A=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]$ and $B=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$ then $A-B$ is equal to
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2626 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} I$
$\frac{1}{2} I$
Since, $A=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1} x \pi & \tan ^{-1} \frac{x}{\pi} \\ \sin ^{-1} \frac{x}{\pi} & \cot ^{-1} \pi x\end{array}\right]$ and $B=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1} x \pi & \tan ^{-1} \frac{x}{\pi} \\ \sin ^{-1} \frac{x}{\pi} & -\tan ^{-1} \pi x\end{array}\right]$
So,
$$
\begin{aligned}
&A-B=\frac{1}{\pi}\left[\begin{array}{ll}
\sin ^{-1} x \pi+\cos ^{-1} x \pi & \tan ^{-1} \frac{x}{\pi}-\tan ^{-1} \frac{x}{\pi} \\
\sin ^{-1} \frac{x}{\pi}-\sin ^{-1} \frac{x}{\pi} & \cot ^{-1} \pi x+\tan ^{-1} \pi x
\end{array}\right] \\
&=\frac{1}{\pi}\left[\begin{array}{cc}
\frac{\pi}{2} & 0 \\
0 & \frac{\pi}{2}
\end{array}\right]=\frac{1}{2}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\frac{1}{2} I
\end{aligned}
$$
So,
$$
\begin{aligned}
&A-B=\frac{1}{\pi}\left[\begin{array}{ll}
\sin ^{-1} x \pi+\cos ^{-1} x \pi & \tan ^{-1} \frac{x}{\pi}-\tan ^{-1} \frac{x}{\pi} \\
\sin ^{-1} \frac{x}{\pi}-\sin ^{-1} \frac{x}{\pi} & \cot ^{-1} \pi x+\tan ^{-1} \pi x
\end{array}\right] \\
&=\frac{1}{\pi}\left[\begin{array}{cc}
\frac{\pi}{2} & 0 \\
0 & \frac{\pi}{2}
\end{array}\right]=\frac{1}{2}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\frac{1}{2} I
\end{aligned}
$$
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