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I'm asking this because I've seen contradictory answers in different textbooks.

On one hand, $\frac{1}{\tan\left(x\right)}$ is undefined for $\tan(x)=0$, but $\tan(x)$ itself is undefined for $x=\frac{\pi}{2}$. So, in this view, the domain should be $\mathbb{R}\backslash \left\{\frac{\pi }{2}n,\:n\in \mathbb{Z}\right\}$.

On the other hand, $\frac{1}{\tan\left(x\right)}$ can also be written as $\frac{\cos\left(x\right)}{\sin\left(x\right)}$, and this is only undefined when $\sin\left(x\right)=0$

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    $\begingroup$ I'd say the second solution, as $\frac1{\tan x}=\cot x$, and the domain of the latter is $\mathbf R\smallsetminus\pi\mathbf Z$. $\endgroup$ – Bernard Mar 24 at 0:31
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    $\begingroup$ The expression $\frac{1}{\tan(x)}$ is computed by first computing the tangent, and the taking the reciprocal of the result. It is then defined when both operations are defined. The expression $\frac{\cos(x)}{\sin(x)}$, although returning the same values at the common points of definition, is a different one. Compare to computing the domain of $\frac{x^2-1}{x-1}$ vs the domain of $x+1$. $\endgroup$ – user647486 Mar 24 at 0:35
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The first answer is correct. $\tan x = \frac{\sin x}{\cos x}$ and so $\frac{1}{\tan x}$ can be written as $\frac{\cos x}{\sin x}$ provided you can actually take the reciprocal of both sides of the original equation. If they are undefined, like when $x = \frac{\pi}{2}$, taking the reciprocal is not a legal operation so the identity breaks down. It is true that $\lim_{x \rightarrow \frac{\pi}{2}}\frac{1}{\tan x}$ is defined and it's equal to $\frac{\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}}$.

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