Solution

1. Let \(x = sin\theta\). Then \(sin^{-1}x = \theta\). We have \(sin^{-1}(2x\sqrt{1-x^2}) = sin^{-1}(2sin\theta\sqrt{1-sin^2\theta})\) \(= sin^{-1}(2sin\theta cos\theta) = sin^{-1}(sin2\theta) = 2\theta\) \(= 2sin^{-1}x\)

2. Take \(x = cos\theta\), then proceeding as above, we get, \(sin^{-1}(2x\sqrt{1-x^2}) = 2cos^{-1}x\)