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Question: \(\sin ^ { - 1 } \left[ x \sqrt { 1 - x } - \sqrt { x } \sqrt { 1 - x ^ { 2 } } \right] =\)...

sin⁡−1[x1−x−x1−x2]=\sin ^ { - 1 } \left[ x \sqrt { 1 - x } - \sqrt { x } \sqrt { 1 - x ^ { 2 } } \right] =

A

sin⁡−1x+sin⁡−1x\sin ^ { - 1 } x + \sin ^ { - 1 } \sqrt { x }

B

sin⁡−1x−sin⁡−1x\sin ^ { - 1 } x - \sin ^ { - 1 } \sqrt { x }

C

sin⁡−1x−sin⁡−1x\sin ^ { - 1 } \sqrt { x } - \sin ^ { - 1 } x

D

None of these

Answer

sin⁡−1x−sin⁡−1x\sin ^ { - 1 } x - \sin ^ { - 1 } \sqrt { x }

Explanation

Solution

Let x=sin⁡θx = \sin \theta and x=sin⁡ϕ\sqrt { x } = \sin \phi

Hence sin⁡−1(x1−x−x1−x2)\sin ^ { - 1 } \left( x \sqrt { 1 - x } - \sqrt { x } \sqrt { 1 - x ^ { 2 } } \right)

=sin⁡−1(sin⁡θ1−sin⁡2ϕ−sin⁡ϕ1−sin⁡2θ)= \sin ^ { - 1 } \left( \sin \theta \sqrt { 1 - \sin ^ { 2 } \phi } - \sin \phi \sqrt { 1 - \sin ^ { 2 } \theta } \right)

=θ−ϕ=sin⁡−1(x)−sin⁡−1(x)= \theta - \phi = \sin ^ { - 1 } ( x ) - \sin ^ { - 1 } ( \sqrt { x } ).