A = frac{1}{pi} begin{bmatrix} sin^{-1}(pi x) | vedclass

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(C) We are given matrices $A$ and $B$. We need to compute $A - B$. $A - B = \frac{1}{\pi} \left[ \begin{bmatrix} \sin^{-1}(\pi x) & 	an^{-1}(\frac{x}

Vedclass · Accepted Answer

$\frac{1}{2} I$

Vedclass · Answer

(C) We are given matrices $A$ and $B$. We need to compute $A - B$. $A - B = \frac{1}{\pi} \left[ \begin{bmatrix} \sin^{-1}(\pi x) & 	an^{-1}(\frac{x}{\pi}) \ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} - \begin{bmatrix} -\cos^{-1}(\pi x) & 	an^{-1}(\frac{x}{\pi}) \ \sin^{-1}(\frac{x}{\pi}) & -	an^{-1}(\pi x) \end{bmatrix} ight]$ Subtracting the corresponding elements: $A - B = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) - (-\cos^{-1}(\pi x)) & 	an^{-1}(\frac{x}{\pi}) - 	an^{-1}(\frac{x}{\pi}) \ \sin^{-1}(\frac{x}{\pi}) - \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) - (-	an^{-1}(\pi x)) \end{bmatrix}$ Using the trigonometric identities $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ and $	an^{-1}(	heta) + \cot^{-1}(	heta) = \frac{\pi}{2}$: $A - B = \frac{1}{\pi} \begin{bmatrix} \frac{\pi}{2} & 0 \ 0 & \frac{\pi}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2} I$ Thus,the correct option is $C$.

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