Some time ago I was wondering about the definition of cotangent; namely, that it is both defined as $\frac{\cos x}{\sin x}$ and as $\frac{1}{\tan x}$. However, $\cot(90°)$ and $\cot(270°)$ are equal to 0. Through the first definition I listed of cotangent, this is equal to $$\cot(90°) =\frac{\cos(90°)}{\sin(90°)} = \frac{0}{1} = 0$$

With the second definition, this is equal to $$\cot(90°) = \frac{1}{\tan(90°)} = \frac{1}{undefined} =\ ?$$

While I know that at $\tan(90°)$ there is an asymptote which, from the right goes to $-\infty$ and from the left goes to $\infty$. While I understand that in this case, the limit $\lim_{x\to \frac{\pi}{2}}\frac{1}{\tan x} = 0$, wouldn't this constitute a domain error in the function $\cot x$ since $\frac{1}{\infty}$ technically doesn't equal $0$?

Thanks for reading.

  • $\begingroup$ It is no different as defining $f(x) = x$ as $\frac{1}{1/x}$. Depends on definition i think. $\endgroup$ – King Tut Feb 1 '18 at 17:31

Cotangent is not $\frac 1\tan$. That is not the definition, and never has been. It just so happens that the two functions $\cot$ and $\frac1\tan$ agree on almost all of $\Bbb R$. When that happens, there is a strict sense in which we consider the functions equivalent, and some authors misuse the symbol $=$ to represent this.

That being said, kudos to you for stopping up and thinking about it.

  • $\begingroup$ Then is the exact definition of cotangent $\frac {\cos x}{\sin x}$? $\endgroup$ – Gil Keidar Feb 1 '18 at 17:37
  • $\begingroup$ @GilKeidar Yes, that's right. $\endgroup$ – Arthur Feb 1 '18 at 17:37

Just $$\cot{x}\neq\frac{1}{\tan{x}}$$ on domain of $\cot.$

We can write $\cot{x}=\frac{1}{\tan{x}}$ for $\sin{x}\cos{x}\neq0$ only.


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