Question

If $A=\frac{1}{\pi }\left[\begin{array}{cc}{\sin }^{-1}(\pi x)& {\tan }^{-1}(\frac{x}{\pi })\\ {\sin }^{-1}\frac{x}{\pi }& {\cot }^{-1}(\pi x)\end{array}\right]$ , $B=\frac{1}{\pi }\left[\begin{array}{cc}-{\cos }^{-1}(\pi x)& {\tan }^{-1}(\frac{x}{\pi })\\ {\sin }^{-1}(\frac{x}{\pi })& -{\tan }^{-1}(\pi x)\end{array}\right]$ then $A-B$ is equal to

• A. $I$
• B. $0$
• C. $2I$
• D. $\frac{1}{2}I$

Answer 1

Answer: (D)

We have, $A=\frac{1}{\pi }\left[\begin{array}{cc}{\sin }^{-1}(\pi x)& {\tan }^{-1}\left(\frac{x}{\pi }\right)\\ {\sin }^{-1}\left(\frac{x}{\pi }\right)& {\cot }^{-1}(\pi x)\end{array}\right]$
and $B=\frac{1}{\pi }\left[\begin{array}{cc}-{\cos }^{-1}(\pi x)& {\tan }^{-1}\left(\frac{x}{\pi }\right)\\ {\sin }^{-1}\left(\frac{x}{\pi }\right)& -{\tan }^{-1}(\pi x)]\end{array}\right]$
$\therefore A-B$
$=\frac{1}{\pi }\left[\begin{array}{cc}{\sin }^{-1}(\pi x)+{\cos }^{-1}(\pi x)& {\tan }^{-1}\left(\frac{x}{\pi }\right)-{\tan }^{-1}\left(\frac{x}{\pi }\right)\\ {\sin }^{-1}\left(\frac{x}{\pi }\right)-{\sin }^{-1}\left(\frac{x}{\pi }\right)& {\cot }^{-1}(\pi x)+{\tan }^{-1}(\pi x)\end{array}\right]$
$=\frac{1}{\pi }\left[\begin{array}{cc}\frac{\pi }{2}& 0\\ 0& \frac{\pi }{2}\end{array}\right]\left[\begin{array}{c}\because {\sin }^{-1}x+{\cos }^{-1}x=\pi /2\\ \text{ and }{\cot }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}\end{array}\right]$
$=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\frac{1}{2}I$