We have to prove that arc sin x + arc cos x = pi/2

The left hand side is

arc sin x + arc cos x

take the sine of the angle

sin (arc sin x + arc cos x)

=> sin (arc sin x)* cos (arc cos x) + cos (arc sin x)* sin (arc cos x)

use the relation sin (arc cos x) = cos (arc sin x) = sqrt (1 - x^2)

=> x*x + [sqrt (1 - x^2)]^2

=> x^2 + 1 - x^2

=> 1

Take the sine of the right hand side

sin pi/2 = 1

We get the same value in both the cases

**This proves that arc sin x + arc cos x = pi/2**

By definition, a function is constant if and only if it's first derivative is cancelling.

We'll assign a function f(x) to the given expression arcsin x + arccos x.

f(x) = arcsin x + arccos x

We'll have to do the first derivative test.

f'(x) = (arcsin x + arccos x)'

f'(x) = [1/sqrt(1-x^2)] - [1/sqrt(1-x^2)]

We'll eliminate like terms:

f'(x)=0,

Since the first derivative was zero, f(x) = constant.

To prove that the constant is pi/2, we'll put x = 1:

**f(1) = arcsin 1 + arccos 1 = pi/2 + 0 = pi/2**