How do you simplify \[\dfrac{{\cos (x + 2\pi )}}{{\sin x}}?\]

Hint: In the given problem first we try to reduce the numerator in the simplest form by eliminating \[2\pi \] such that we are able to simplify it by using trigonometric formula.
Formula used:
i). \[\cos (2\pi + x) = \cos x\]
ii). \[\dfrac{{\cos x}}{{\sin x}} = \cot x\]

Complete step-by-step solution:
First writing the given expression as follows,
\[ = \dfrac{{\cos (x + 2\pi )}}{{\sin x}}\]
Arranging numerator and writing it as following,
\[ = \dfrac{{\cos (2\pi + x)}}{{\sin x}}\]
By using above given formula, we can write numerator as following,
\[ = \dfrac{{\cos x}}{{\sin x}}\]
Again, by applying above given trigonometric formula, we get
 \[ = \dfrac{{\cos x}}{{\sin x}}\]
\[ = \cot x\]

Note: We have to keep in mind that when we add \[2\pi \] or its multiple to any trigonometric ratio does not change its value. For example,
\[\sin (x + 2n\pi ) = \sin x\] where, \[n = 0,1,2,3,4,.....,\infty \]
\[\cos (x + 2n\pi ) = \cos x\] where, \[n = 0,1,2,3,4,.....,\infty \]
\[\tan (x + 2n\pi ) = \tan x\] where, \[n = 0,1,2,3,4,.....,\infty \]
\[\cos ec(x + 2n\pi ) = \cos ecx\] where, \[n = 0,1,2,3,4,.....,\infty \]
\[\sec (x + 2n\pi ) = \sec x\] where, \[n = 0,1,2,3,4,.....,\infty \]
\[\cot (x + 2n\pi ) = \cot x\] where, \[n = 0,1,2,3,4,.....,\infty \]
For the above used directly trigonometric formula can be understood by using right-angle triangle and naming its sides as perpendicular, base, hypotenuse. In a right-angle triangle in which one angle is 90 degree and the side which exists in front of that 90-degree is named as hypotenuse which is also the longest side of the triangle. Rest of the two sides are named as base and perpendicular.
Reduction of \[\dfrac{{\cos x}}{{\sin x}}\] to \[\cot x\] as following,
\[\dfrac{{\cos x}}{{\sin x}}\] can be written in the form of perpendicular, base and hypotenuse as following,
\[ = \]\[\dfrac{{\cos x}}{{\sin x}}\]
\[\cos x\] can be written as following,
\[ = \]\[\cos x\]
\[ = \]\[\dfrac{{base}}{{hypotenuse}}\]
Or \[\dfrac{b}{h}\] where, (b=base, h=hypotenuse)
\[\sin x\] can be written as following,
\[ = \]\[\sin x\]
\[ = \]\[\dfrac{{perpendicular}}{{hypotenuse}}\]
Or \[\dfrac{p}{h}\] where, (p=perpendicular, h=hypotenuse)
Now, \[\dfrac{{\cos x}}{{\sin x}}\] can be written as following,
\[ = \dfrac{{\cos x}}{{\sin x}}\]
Replacing \[\cos x\] by (b=base, h=hypotenuse) and writing it as,
\[ = \dfrac{{\dfrac{b}{h}}}{{\sin x}}\]
Replacing \[\sin x\] by (p=perpendicular, h=hypotenuse) and writing it as,
\[ = \dfrac{{\dfrac{b}{h}}}{{\dfrac{p}{h}}}\]
Cancelling denominator and writing it as,
\[ = \dfrac{b}{p}\]
Now we know that \[\cot x\] is equal to \[\dfrac{b}{p}\].
Therefore, \[\dfrac{b}{p}\] can be also written as following,
\[ \Rightarrow \dfrac{b}{p} = \cot x\]
Hence, \[\dfrac{{\cos x}}{{\sin x}}\] can be written as \[\cot x\].
Therefore, \[\dfrac{{\cos x}}{{\sin x}} = \]\[\cot x\].