For both sides to have sense, x must be in [-1, 1].
For such x'es, `sin^(-1)(x)` is the angle y in `[-pi/2, pi/2]` such that `sin(y) = x.`
But `cos^2(y) + sin^2(y) = 1,` so `cos^2(y) = 1 - sin^2(y) = 1-x^2.`
Also, for y in `[-pi/2, pi/2]` always cos(y)>=0. Therefore we can state that
`cos(y) = cos(sin^(-1)(x)) = sqrt(1-x^2)` (not `-sqrt(1-x^2)` ), QED.
Taking x as opposite sid and 1 as hypotenuse we get adjacent side as `sqrt(1-x^2).`
`cos(sin^-1x) =cos(cos^-1(sqrt(1-x^2)/1)) = sqrt(1-x^2).`