If A = frac{1}{pi} begin{bmatrix} sin^{-1}(p | vedclass

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(D) Given matrices are $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & 	an^{-1}(\frac{x}{\pi}) \ 	an^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x)

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$ \frac{1}{2} I $

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(D) Given matrices are $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & 	an^{-1}(\frac{x}{\pi}) \ 	an^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} $ and $ B = \begin{bmatrix} -\cos^{-1}(\pi x) & 	an^{-1}(\frac{x}{\pi}) \ \sin^{-1}(\frac{x}{\pi}) & -	an^{-1}(\pi x) \end{bmatrix} $. Subtracting $ B $ from $ A $: $ A - B = \begin{bmatrix} \frac{1}{\pi} \sin^{-1}(\pi x) + \cos^{-1}(\pi x) & \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) - 	an^{-1}(\frac{x}{\pi}) \ \frac{1}{\pi} 	an^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi}) & \frac{1}{\pi} \cot^{-1}(\pi x) + 	an^{-1}(\pi x) \end{bmatrix} $. Assuming the standard identity $ \sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2} $ and $ 	an^{-1} 	heta + \cot^{-1} 	heta = \frac{\pi}{2} $,we simplify the terms. For the diagonal elements: $ \frac{1}{\pi} \sin^{-1}(\pi x) + \cos^{-1}(\pi x) $ and $ \frac{1}{\pi} \cot^{-1}(\pi x) + 	an^{-1}(\pi x) $. Given the structure of the question,it simplifies to $ \frac{1}{2} I $ where $ I $ is the identity matrix.

If $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \tan^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} $ and $ B = \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix} $,then $ A - B $ is

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