arcsin x + arccosx = pi/2 if -1<x<1

Let f(x) = arcsinx + arccosx

If f(x) = 1 , then it is a constant , then f'(x) should be 0:

f'(x) = (arcsinx + arccosx)'

= 1/sqrt(1-x^2) -1/sqrt(1-x^2) = 0

Then we proved that f(x) is a constant number

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arcsin x + arccosx = pi/2 if -1<x<1

Let f(x) = arcsinx + arccosx

If f(x) = 1 , then it is a constant , then f'(x) should be 0:

f'(x) = (arcsinx + arccosx)'

= 1/sqrt(1-x^2) -1/sqrt(1-x^2) = 0

Then we proved that f(x) is a constant number

==> f(x) = C

Now let us substitutw with any value within the interval given.

f(1) = arcsin1 + arccos1 = pi/2 + 0 = pi/2

==> arcsinx + arccosx = pi/2