Both solutions are correct, because
[tex]\arcsin(x/a) = \arctan(x/\sqrt{a^2-x^2})[/tex].
You can prove this by considering a right angled triangle with hypotenuse [tex]a[/tex], opposite side [tex]x[/tex], and by Pythagoras the adjacent side is [tex]\sqrt{a^2-x^2}[/tex]. So [tex]\sin\theta = x/a[/tex], and [tex]\tan\theta = x/\sqrt{a^2-x^2}[/tex], which means we have two ways of calculating [tex]\theta[/tex] which results in [tex]\theta = \arcsin(x/a) = \arctan(x/\sqrt{a^2-x^2})[/tex].
See also
https://en.wikipedia.org/wiki/Inverse_t ... _functionsHope this helped,
R. Baber.