Yes, the problem could be solved using calulus methods.
For instance, if we'll consider the expression (arcsin x + arccos x) as a function f(x), in order to demonstrate that f(x)=pi/2 is a constant function, we'll have to calculate the first derivative of this function f(x).
If this derivative is cancelling, that means that f(x)=pi/2, is a constant function, knowing the fact that a derivative of a constant function is 0.
We'll take the first derivative of the given function:
f'(x) = (arcsin x + arccos x)'
f'(x) = 1/sqrt(1-x^2) - 1/sqrt(1-x^2)
f'(x)=0 => so f(x)=constant
We notice that for x = 1, the sum of inverse trigonometric functions is pi/2.
f(1)=arcsin 1 + arccos 1 = pi/2 + 0=pi/2