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I believe $\cot x$ should be defined as $ \dfrac{\cos x}{\sin x}$ and not as $\dfrac{1}{\tan x}$

Because $\cot x$ and $\dfrac{1}{\tan x}$ aren't even the same function, they have different domains.

So for instance we know $\cot (π/2) = 0$ but $\dfrac{1}{ \tan (π/2)}$ is not even defined.

Am I correct in believing this?

William
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    You are right about this. – José Carlos Santos May 27 '19 at 14:35
  • @JoséCarlosSantos Ty, most of the books, in fact all the sources I could find define $\cot x$ as reciprocal of $\tan x$. Wonder why. – William May 27 '19 at 14:37
  • @N.F.Taussig: I was going to say something similar, though I think you meant cot$\theta=\frac {\cos\theta}{\sin\theta}$ – J. W. Tanner May 27 '19 at 14:54
  • @J.W.Tanner Well, that was an embarrassing error. – N. F. Taussig May 27 '19 at 14:59
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    I suspect the issue is that the authors of the texts that define $\cot\theta$ as the reciprocal of $\tan\theta$ are doing so for trigonometric ratios rather than trigonometric functions. Since trigonometric ratios are defined for acute angles and it is not possible for $\tan\theta = 0$ when $\theta$ is acute, such a definition does not present a problem until you consider trigonometric functions. For trigonometric functions $f(x) = \cot x = \frac{\cos x}{\sin x}$ is defined whenever $\sin x \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$. – N. F. Taussig May 27 '19 at 15:03
  • by analytic continuation they have the same domain, I think this is the reason to state both as equivalent expressions (under analytic continuation considerations, of course) – Masacroso May 27 '19 at 15:19

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