For both sides to have sense, x must be in [-1, 1]. For those x'es `sin^(-1)(x)` is the angle y in `[-pi/2, pi/2]` such that sin(y)=x.
cos is nonnegative on `[-pi/2, pi/2],` so `cos(y) = sqrt(1-sin^2(y)) = sqrt(1-x^2).` And
`tan(sin^(-1)(x)) = tan(y) = sin(y)/cos(y) = x/sqrt(1-x^2),` QED.
Taking x as opposite side and 1 as hypotenuse we get adjacent side as `sqrt(1-x^2).`
`tan(sin^-1x) = tan(tan^-1(x/(sqrt(1-x^2)))) = x/sqrt(1-x^2)`