How do you simplify $\frac{cos (x + 2 π)}{sin x}$ $[cos(x+2pi)]/sinx$ ?

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How do you simplify $\frac{cos (x + 2 π)}{sin x}$ $[cos(x+2pi)]/sinx$ ?

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Answer 1 · 2016-02-22T05:41:38

$\frac{cos (x + 2 π)}{sin x}$ $cos(x+2pi)/sinx$ = $cot x$ $cotx$

Explanation:

To simplify $\frac{cos (x + 2 π)}{sin x}$ $cos(x+2pi)/sinx$ , first one may note that adding $2 π$ $2pi$ or its multiple, to a trigonometric ratio, does not change its value

i.e. $sin (x + 2 n π) = sin x$ $sin(x+2npi)=sinx$ (for all $n$ $n$ , where $n$ $n$ is an integer), $cos (x + 2 n π) = cos x$ $cos(x+2npi)=cosx$ , $tan (x + 2 n π) = tan x$ $tan(x+2npi)=tanx$ , $cot (x + 2 n π) = cot x$ $cot(x+2npi)=cotx$ , $sec (x + 2 n π) = sec x$ $sec(x+2npi)=secx$ and $csc (x + 2 n π) = csc x$ $csc(x+2npi)=cscx$ .

Hence, $\frac{cos (x + 2 π)}{sin x}$ $cos(x+2pi)/sinx$ = $\frac{cos x}{sin x}$ $cosx/sinx$ or $cot x$ $cotx$ .

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How do you simplify $\frac{cos (x + 2 π)}{sin x}$ $[cos(x+2pi)]/sinx$ ?

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How do you simplify $\frac{cos (x + 2 π)}{sin x}$ $[cos(x+2pi)]/sinx$ ?