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# If $\;A=\small\frac{1}{\pi}$$\begin{bmatrix}sin^{-1}(\pi x) &tan^{-1}\big(\frac{\pi}{x}\big)\\ sin^{-1}\big(\frac{\pi}{x} \big)&cot^{-1}(\pi x)\end{bmatrix}\; and \;B=\small\frac{1}{\pi}$$\begin{bmatrix}-cos^{-1}(\pi x) &tan^{-1}\big(\frac{x}{\pi}\big)\\ sin^{-1}\big(\frac{x}{\pi} \big)&-tan^{-1}(\pi x)\end{bmatrix}$ then $A-B$ is equal to

$(A)\;I \quad(B)\;0\quad(C)\;2I\quad(D)\;\frac{1}{2} I$

Toolbox:
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1 $\leq$j $\leq$ n.
• $sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$
• $cot ^{-1}x+tan^{-1}x=\frac{\pi}{2}$
Step1:
Given
$A= \frac{1}{\pi}\begin{bmatrix}sin^{-1}(\pi x) & tan^{-1}(x/\pi)\\sin^{-1}(\pi/x) & cot ^{-1}(\pi x)\end{bmatrix}$
$B=\frac{1}{\pi}\begin{bmatrix}-cos^{-1}(\pi x)& tan^{-1}(x/\pi)\\sin^{-1}(\pi/x)& -tan^{-1}(\pi x)\end{bmatrix}$
To find: $A-B=A+(-1)B$, Substitute the value of $A$ and $B$:
$\Rightarrow \frac{1}{\pi}\begin{bmatrix}sin^{-1}(\pi x) & tan^{-1}(x/\pi)\\sin^{-1}(\pi/x) & cot ^{-1}(\pi x)\end{bmatrix}+\frac{1}{\pi}(-1)\begin{bmatrix}-cos^{-1}(\pi x)& tan^{-1}(x/\pi)\\sin^{-1}(\pi/x)& -tan^{-1}(\pi x)\end{bmatrix}$
$\Rightarrow \frac{1}{\pi}\begin{bmatrix}sin^{-1}(\pi x)+cos^{-1}(\pi x) & tan^{-1}(x/\pi)-tan^{-1}(x/\pi)\\sin^{-1}(\pi/x)-sin^{-1}(\pi/x) & cot ^{-1}(\pi x)+tan^{-1}(\pi x)\end{bmatrix}$
Step2:
We know that
$sin^{-1}x+cos^{-1}x=\frac{\pi}{2}$
$cot ^{-1}x+tan^{-1}x=\frac{\pi}{2}$
$\Rightarrow \frac{1}{\pi}\begin{bmatrix}\pi/2 & 0\\0 & \pi/2\end{bmatrix}$
$\Rightarrow \frac{1}{\pi}.\frac{\pi}{2}\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\qquad=\frac{1}{2}I$
edited Mar 19, 2014