Verify if arctanx=pi/2-arccotx. -1

The problem could be solved using calulus
methods.

We'll create a function f(x) = arctan x + arccot
x.

We should prove that f(x) = arctan x + arccot x =
pi/2.

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 out, 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 differentiate to find out the first
derivative of the given function:

f'(x) = (arctan x +
arccot x)'

f'(x) = 1/(1+x^2) -
1/(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)=arctan 1 + arccot 1 =
pi/4 + pi/4 = 2pi/4 = pi/2

The identity
arctanx=pi/2-arccotx is verified for x = 1.