To prove arctanx +arc cotx = pi/2. T

Let arc cot x = a

x = cot a.

1/x = tana.

Therefore arc (1/x) = a = arc cot x

Therefore , arc tanx +arc cotx = arc tanx +arc an (1/x).

To find arctanx +arctan (1/x) = y say

tany = (x+1/x)/(1-x*1/x) = (x+1/x)/ 0.

Therefore tany = infinity.

Therefore y = pi/2. Or

arctanx +arc cotx = pi/2.

arctanx + arccotx = pi/2

We'll associate a function f(x) to the expression (arctanx + arccotx).

If we have to verify if the function is a constant function, we'll have to do the first derivative test.

When the first derivative of a function is cancelling, that means that the function is a constant function, because the derivative of a constant function is 0.

We'll differentiate the function f(x):

f'(x) = (arctanx + arccotx)'

f'(x) = 1/(1+x^2) - 1/(1+x^2)

We'll eliminate like terms:

f'(x)=0,

If f'(x)=0 => f(x)=constant

Now, we'll determine the constant for (arctanx + arccotx )' = 0

We'll prove that the constant is pi/2.

For this reason, we'll put x = 1:

f(1)=arctan1+ arccot1= pi/2

f(1) = pi/4 + pi/4

f(1) = 2pi/4

**f(1) = pi/2 q.e.d.**