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  • If A = 1/pi [sin inverse x pi]. B = 1/pi[-cos inverse (xpi)] then A - B is equal to A. I

If A=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & \cot ^{-1}(\pi \mathrm{x}) \end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & -\tan ^{-1}(\pi \mathrm{x}) \end{array}\right] then A - B is equal to
A. I
B. O
C. 2I
D. \frac{1}{2}I

Answers (1)

We will use Inverse trigonometric function to solve the problem

cos\textsuperscript{-1} x + sin\textsuperscript{-1} x = \pi /2 \: \: and \: \: cot\textsuperscript{-1} x + tan\textsuperscript{-1} x = \pi /2

As A=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & \cot ^{-1}(\pi \mathrm{x}) \end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & -\tan ^{-1}(\pi \mathrm{x}) \end{array}\right]

\begin{array}{l} \therefore A-B=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(x \pi)+\cos ^{-1}(x \pi) & 0 \\ 0 & \tan ^{-1} \pi x+\cot ^{-1} \pi x \end{array}\right] \\ \Rightarrow A-B=\left[\begin{array}{cc} \frac{\pi}{2} \times \frac{1}{\pi} & 0 \\ 0 & \frac{\pi}{2} \times \frac{1}{\pi} \end{array}\right]=\left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \\ \therefore A-B=\frac{1}{2}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{2} I \end{array}

As it matches with option (D)

Hence, we can say that,

∴ option(D) is the only correct answer.

Posted by

infoexpert22

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