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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.
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