Answer:
Consider \[A\,-B=\frac{1}{\pi }\left\{ \left[ \begin{matrix} {{\sin }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right) \\ {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & {{\cot }^{-1}}(x\pi ) \\ \end{matrix} \right] \right.\] \[\left. -\left[ \begin{matrix} -{{\cos }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right) \\ {{\sin }^{-1}}\left( \frac{x}{\pi } \right) & -{{\tan }^{-1}}(\pi x) \\ \end{matrix} \right] \right\}\] \[=\frac{1}{\pi }\left[ \begin{matrix} {{\sin }^{-1}}(x\pi )+{{\cos }^{-1}}(x\pi ) & {{\tan }^{-1}}\left( \frac{x}{\pi } \right)-{{\tan }^{-1}}\left( \frac{x}{\pi } \right) \\ {{\sin }^{-1}}\left( \frac{x}{\pi } \right)-{{\sin }^{-1}}\left( \frac{x}{\pi } \right) & {{\cot }^{-1}}(x\pi )+{{\tan }^{-1}}(\pi x) \\ \end{matrix} \right]\] \[=\frac{1}{\pi }\left[ \begin{matrix} \frac{\pi }{2} & 0 \\ 0 & \frac{\pi }{2} \\ \end{matrix} \right]\] \[\left[ \begin{align} & \because {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2};x\in []-1,1 \\ & \text{and}\,\,{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2},x\in R \\ \end{align} \right]\] \[=\left[ \begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \\ \end{matrix} \right]=\frac{1}{2}l\]
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