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प्रश्न
If \[A = \frac{1}{\pi}\begin{bmatrix}\sin^{- 1} \left( \ pix \right) & \ tan^{- 1} \left( \frac{x}{\pi} \right) \\ \sin^{- 1} \left( \frac{x}{\pi} \right) & \cot^{- 1} \left( \ pix \right)\end{bmatrix}, B = \frac{1}{\pi}\begin{bmatrix}- \cos^{- 1} \left( \ pix \right) & \tan^{- 1} \left( \frac{x}{\pi} \right) \\ \sin^{- 1} \left( \frac{x}{\pi} \right) & - \tan^{- 1} \left( \ pix \right)\end{bmatrix}\]
A − B is equal to
विकल्प
I
0
2I
`1/2 I`
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उत्तर
Given:
\[A - B = \frac{1}{\pi}\begin{bmatrix}\sin^{- 1} \left( \pi x \right) & \tan^{- 1} \left( \frac{x}{\pi} \right) \\ \sin^{- 1} \left( \frac{x}{\pi} \right) & \cot^{- 1} \left( \pi x \right)\end{bmatrix} - \frac{1}{\pi}\begin{bmatrix}- \cos^{- 1} \left( \pi x \right) & \tan^{- 1} \left( \frac{x}{\pi} \right) \\ \sin^{- 1} \left( \frac{x}{\pi} \right) & - \tan^{- 1} \left( \pi x \right)\end{bmatrix}\]
\[ = \frac{1}{\pi}\left( \begin{bmatrix}\sin^{- 1} \left( \pi x \right) & \tan^{- 1} \left( \frac{x}{\pi} \right) \\ \ sin^{- 1} \left( \frac{x}{\pi} \right) & \ cot^{- 1} \left( \pi x \right)\end{bmatrix} + \begin{bmatrix}\cos^{- 1} \left( \pi x \right) & - \tan^{- 1} \left( \frac{x}{\pi} \right) \\ - \sin^{- 1} \left( \frac{x}{\pi} \right) & \tan^{- 1} \left( \pi x \right)\end{bmatrix} \right)\]
\[ = \frac{1}{\pi}\begin{bmatrix}\sin^{- 1} \left( \pi x \right) + \cos^{- 1} \left( \pi x \right) & \tan^{- 1} \left( \frac{x}{\pi} \right) - \tan^{- 1} \left( \frac{x}{\pi} \right) \\ \sin^{- 1} \left( \frac{x}{\pi} \right) - \sin^{- 1} \left( \frac{x}{\pi} \right) \& \cot^{- 1} \left( \pi x \right) + \tan^{- 1} \left( \pi x \right)\end{bmatrix}\]
\[ = \frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2}\end{bmatrix} \left( \because \sin^{- 1} x + \cos^{- 1} x = \frac{\pi}{2} \text{ and } \cot^{- 1} x + \tan^{- 1} x = \frac{\pi}{2} \right)\]
\[ = \frac{1}{\pi} \times \frac{\pi}{2}\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ = \frac{1}{2}I\]
Hence, the correct option is (d).
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