(i)sin⁡−1(1−x)−2sin⁡−1x=π2(i) \sin^{-1}(1-x) - 2 \sin^{-1} x = \frac{\pi}{2}

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(i)sin⁡−1(1−x)−2sin⁡−1x=π2(i) \sin^{-1}(1-x) - 2 \sin^{-1} x = \frac{\pi}{2}

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