What is the value of x if cos^-1 x = 90-sin^-1 x?

We'll re-write the identity, using
arcfunctions:

arccos x = pi/2 - arcsin
x

We'll shift arcsin x to the left side and we'll
get:

arccos x + arcsin x =
pi/2

It is obvious that for x = 1, we'll
get:

arcsin 1 + arccos 1 = pi/2 + 0 =
pi/2

We have to prove that x = 1 is the only
solution.

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

f(x) = arcsin x + arccos
x

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

A continuous
differentiable function is also bijective, therefore it is injective, so x = 1 is the
only solution for the equation arccos x = pi/2 - arcsin
x.