Hello Aspirant,

I think there domain will be -1<x<1

Since value in sin-1 can only be between -1 and 1, we get the conditions

1. -1 < x < 1
2. -1/2 < x√1-x2 < 1/2

Taking case 2, square it. We get x2(1-x2) <1/4

Replace x2 with t. We get t2-t+1/4>0.

The LHS is a square , thus is greater than zero for all t . Therefore for all x .

The intersection of the two ranges gives -1<x<1

I hope this will help you.