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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:
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.
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