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I was thinking about trigonometry ratios, in particularly $\cot(\theta)$, which can be defined as $\cot(\theta) = \frac {1}{\tan(\theta)} = \frac {cos(\theta)}{sin(\theta)}$. Though $\tan(90)$ is not defined as you end up getting $\frac{1}{0}$. Though $\cot(90) = 0$. Though one could interpret that we had to divide it by something undefined as $\frac {1}{0}$, isn't defined. Yet divided by it and we get something defined.

Is it ok to do this?

Is it always ok to divide by something undefined, if it could be flipped in such a way that you get something defined?

Thanks

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  • $\begingroup$ The formula you give as "$\cot\theta=1/\tan\theta$" is more correctly given as "$\cot\theta=1/\tan\theta$ provided $\theta$ is not a multiple of $\pi/2$." $\endgroup$ May 27 '16 at 9:10
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You may just define $\cot(x) = \frac{\cos(x)}{\sin(x)}$. Then you do not run into problems when evaluating $\cot(\frac\pi2)$.

Without somehow giving a meaning to the term $\frac10$, the equality $\cot(x) = \frac{1}{\tan(x)}$ holds only for those $x$ where $\tan(x)$ is not equal to zero.

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  • $\begingroup$ Ok, but in general can I divide by something undefined? Is $\frac {1}{\frac {1}{0}}$ defined? Is it 0? $\endgroup$
    – frog1944
    May 28 '16 at 1:54
  • $\begingroup$ 1/0 is not defined, at least not as a real number. You can add new symbols to the real numbers, which you may call $\infty$ and $-\infty$ and you may define arithmetic rules for them. But this is not very common. $\endgroup$
    – skew41
    May 28 '16 at 19:10
  • $\begingroup$ Ok, though can you divide by it? $\endgroup$
    – frog1944
    May 29 '16 at 8:52
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    $\begingroup$ To answer if you "can" divide by 0, you must make this more precise. In the real numbers for instance you cannot. $\endgroup$
    – skew41
    May 30 '16 at 14:28

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