Are $\mathrm{arccot}(x)$ 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 () 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:
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$\begingroup$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.
$\endgroup$ 1 $\begingroup$This is not a rigorous proof, it is intuitive one.
Let $y = \arctan\left(\dfrac1x\right)$ then $$ \begin{align} \tan(y) &= \frac1x\\ \frac1{\tan(y)} &= x\\ \cot(y) &= x\\ y &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) - \text{arccot}(x)&=0. \end{align} $$
It holds for $x \neq 0, \ x\in\mathbb{R}$.
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