Identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)

[MathewsMD]

I've been looking for this identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)

After just visually inspecting this to be true, I have been unable to find any formal proofs for it.
Any references would be great!

 

[Chestermiller]

I've been looking for this identity: arctan(1/x) = arcot(x) or arccot(1/x) = arctan(x)

After just visually inspecting this to be true, I have been unable to find any formal proofs for it.
Any references would be great!

The cotangent function is defined as the reciprocal of the tangent function.

 

[gopher_p]

The identity is true for positive values of $x$ but not for negative.

I would suggest starting from the definitions of those functions; $$y=\arctan x\iff \tan y=x \text{ and }y\in (-\frac{\pi}{2},\frac{\pi}{2})$$ and $$y=\text{arc}\cot x\iff \cot y=x \text{ and }y\in (0,\pi)$$ along with the trig identity suggested by Chestermiller.

 

[Chestermiller]

The identity is true for positive values of $x$ but not for negative.

I would suggest starting from the definitions of those functions; $$y=\arctan x\iff \tan y=x \text{ and }y\in (-\frac{\pi}{2},\frac{\pi}{2})$$ and $$y=\text{arc}\cot x\iff \cot y=x \text{ and }y\in (0,\pi)$$ along with the trig identity suggested by Chestermiller.

Is there a reason cot and tan defined on different ranges?

 

[HallsofIvy]

Yes. Since tangent and arctangent are periodic functions, in order to talk about "well-defined" inverses (since $tan(\pi/4)= tan(5\pi/4)= 1$ should arctan(1) be $\pi/4$ or $5\pi/4$?), we have to restrict the range of tangent and cotangent and so restrict the domain of arctangent and arccotangent. The usual convention is to restrict to the largest interval containing 0 on which the function is one-to-one. For the tangent that is $-\pi/2$ to $\pi/2$. Since $cot(x)= tan(\pi/2- x)$, the corresponding domain for cotangent is 0 to $\pi$.