dream13rxs

2022-07-05

Non Permissible values of $\mathrm{cot}\left(x\right)$
Why is it that the non-permissible values of cotangent x is only where $\mathrm{sin}\left(x\right)=0$ and not also where $\mathrm{cos}\left(x\right)=0$

Nicolas Calhoun

Expert

one way you can see this is that
$\mathrm{cot}x=\frac{1}{\mathrm{tan}x}=\frac{1}{\frac{\mathrm{sin}x}{\mathrm{cos}x}}=\frac{\mathrm{cos}x}{\mathrm{sin}x}$
which is 0 when $\mathrm{cos}x=0$, and undefined when $\mathrm{sin}x=0$

Lillianna Andersen

Expert

The tangent of x is defined as the the sine of x divided by the cosine of x, so
$\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$
If x=0 then product is 0/1=0, which is a real solution.
However for
$\mathrm{cot}x=\frac{\mathrm{cos}x}{\mathrm{sin}x}$
Now, if x=0, then the product is 1/0 which is undefined, meaning that a solution does not exist, which means that sinx cannot be equal to 0 for cotx to exist

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