dream13rxs

Answered

2022-07-05

Non Permissible values of $\mathrm{cot}(x)$

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

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

Answer & Explanation

Nicolas Calhoun

Expert

2022-07-06Added 15 answers

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 $\mathrm{sin}x=0$

$\mathrm{cot}x=\frac{1}{\mathrm{tan}x}=\frac{1}{\frac{\mathrm{sin}x}{\mathrm{cos}x}}=\frac{\mathrm{cos}x}{\mathrm{sin}x}$

which is 0 when $\mathrm{cos}x=0$, and undefined when $\mathrm{sin}x=0$

Lillianna Andersen

Expert

2022-07-07Added 3 answers

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

$\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$

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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