# Are arccot(x) and arctan(1/x) the same function?

In my textbook it asks for me to:

Prove that there is no constant $C$ such that $\text{arccot}(x) - \text{arctan}(\frac{1}{x}) = C$ for all $x \ne 0$. Explain why this does not violate the zero-derivative theorem.

But I believe I have found such a $C$, i.e. $C =0$! I even asked WolframAlpha (http://www.wolframalpha.com/input/?i=arccot%28x%29+-+arctan%281%2Fx%29) which corroborates my answer.

This question appears in Apostol's Calculus Volume I, Second Edition: Exersize 6.22-11b

Edit: Mathematica's definition of arccot is different from the one in my textbook. Apostol's arccot maps a real number into $(0, \pi)$ while Mathematica's maps a real number into $(-\pi/2, \pi/2)$ Here they are super-imposed: http://www.wolframalpha.com/input/?i=integral%28-1%2F%281%2Bx%5E2%29%29+%2B+pi%2F2%3B+arccot

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The inverse trig functions are multivalued and therefore not uniquely defined, unless a principle value is given, so please clarify how you are defining the two functions. –  Ethan Feb 14 '13 at 23:25
@Ethan Usually $\arctan$ denotes the determination which takes values in $(-\pi/2,\pi/2)$, no? But you're right, this is not so clear for arccot. –  julien Feb 14 '13 at 23:31
Here: intmath.com/blog/which-is-the-correct-graph-of-arccot-x/6009 there are two definitions of arccot. Which one are you using? –  julien Feb 14 '13 at 23:32
Oh! I had assumed that mathematicians were in agreement. My textbook uses arccot = pi/2 - arctan, which is a different version than Mathematica's according to @julien's link. Thanks! –  Mark Feb 14 '13 at 23:39
It seems that, at least as Wolfram is graphing it, the reflection about $y = x$ is used...(analytic?)...Even though Wolfram evaluates the difference as 0, and returns "true" to the equality of each expression, it nonetheless produces its indefinite integral as $x(\arccot(x) - \arctan(1/x)) + C, which makes little sense, to have done if the difference is zero: why not return just "C"? – amWhy Feb 14 '13 at 23:39 show 3 more comments ## 1 Answer So we'll take your textbook's definition: $$\mbox{arccot}(x)=\frac{\pi}{2}-\arctan(x).$$ Then $$\lim_{0^+} \mbox{arccot} (x)=\frac{\pi}{2}=\lim_{0^+} \arctan(1/x)$$ while $$\lim_{0^-} \mbox{arccot} (x)=\frac{\pi}{2}=-\lim_{0^-} \arctan(1/x).$$ Since its derivative is$0$on$\mathbb{R}^*$, $$\mbox{arccot}(x)-\arctan(1/x)=0$$ for all$x>0$while $$\mbox{arccot}(x)-\arctan(1/x)=\pi$$ for all$x<0$. This does not contradict the zero-derivative theorem because the function is not defined at$0\$, so its domain is not connected.

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And actually, arctan(1/x) gives the alternative definition of arccot(x). –  Mark Feb 15 '13 at 0:09