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

We have, $A = \left[ {\begin{array}{llllllllllllllllllll}{\frac{1}{\pi }{{\sin }^{ - 1}}x\pi }&{\frac{1}{\pi }{{\tan }^{ - 1}}\frac{x}{\pi }}\\{\frac{1}{\pi }{{\sin }^{ - 1}}\frac{x}{\pi }}&{\frac{1}{\pi }{{\cot }^{ - 1}}\pi x}\end{array}} \right]$

and $\quad B = \left[ {\begin{array}{llllllllllllllllllll}{\frac{{ - 1}}{\pi }{{\cos }^{ - 1}}x\pi }&{\frac{1}{\pi }{{\tan }^{ - 1}}\frac{x}{\pi }}\\{\frac{1}{\pi }{{\sin }^{ - 1}}\frac{x}{\pi }}&{\frac{{ - 1}}{\pi }{{\tan }^{ - 1}}\pi x}\end{array}} \right]$

$\therefore$ $\quad A - B = \left[ {\begin{array}{llllllllllllllllllll}{\frac{1}{\pi }\left( {{{\sin }^{ - 1}}x\pi + {{\cos }^{ - 1}}x\pi } \right)}&{\frac{1}{\pi }\left( {{{\tan }^{ - 1}}\frac{x}{\pi } - {{\tan }^{ - 1}}\frac{x}{\pi }} \right)}\\{\frac{1}{\pi }\left( {{{\sin }^{ - 1}}\frac{x}{\pi } - {{\sin }^{ - 1}}\frac{x}{\pi }} \right)}&{\frac{1}{\pi }{{\cot }^{ - 1}}\pi x + {{\tan }^{ - 1}}\pi x}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\frac{1}{\pi }}&{\frac{\pi }{2}}&0\\0&{\frac{1}{\pi }}&{\frac{\pi }{2}}\end{array}} \right]$

$= \frac{1}{2}\left[ {\begin{array}{llllllllllllllllllll}1&0\\0&1\end{array}} \right]$

$= \frac{1}{2}{\rm{I}}$