How do you simplify $(\frac{sin x}{1 - cos x}) + (\frac{1 - cos x}{sin x})$ $(sinx/(1-cosx))+((1-cosx)/sinx)$ ?

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How do you simplify $(\frac{sin x}{1 - cos x}) + (\frac{1 - cos x}{sin x})$ $(sinx/(1-cosx))+((1-cosx)/sinx)$ ?

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Answer 1 · 2016-03-01T23:38:27

$\frac{sin x}{1 - cos x} + \frac{1 - cos x}{sin x} = 2 csc x$ $sinx/(1-cosx)+(1-cosx)/sinx = 2cscx$

Explanation:

$\frac{sin x}{1 - cos x} + \frac{1 - cos x}{sin x}$ $sinx/(1-cosx)+(1-cosx)/sinx$

Multiply the first term by $\frac{sin x}{sin x}$ $sinx/sinx$ and the second term by $\frac{1 - cos x}{1 - cos x}$ $(1-cosx)/(1-cosx)$

$= \frac{{sin}^{2} x}{sin x (1 - cos x)} + \frac{{(1 - cos x)}^{2}}{sin x (1 - cos x)}$ $= sin^2x/(sinx(1-cosx))+(1-cosx)^2/(sinx(1-cosx))$

Group terms with common denominators

$= \frac{{sin}^{2} x + {(1 - cos x)}^{2}}{sin x (1 - cos x)}$ $=(sin^2x+(1-cosx)^2)/(sinx(1-cosx))$

Expand ${(1 - cos x)}^{2}$ $(1-cosx)^2$

$= \frac{{sin}^{2} x + 1 - 2 cos x + {cos}^{2} x}{sin x (1 - cos x)}$ $=(sin^2x+1-2cosx+cos^2x)/(sinx(1-cosx))$

Apply the identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$= \frac{2 - 2 cos x}{sin x (1 - cos x)}$ $=(2-2cosx)/(sinx(1-cosx))$

Factor out $2$ $2$ from the numerator

$= \frac{2 (1 - cos x)}{sin x (1 - cos x)}$ $=(2(1-cosx))/(sinx(1-cosx))$

Cancel common terms from the numerator and denominator

$= \frac{2}{sin x}$ $=2/sinx$

Apply the definition of the cosecant function ( $csc x = \frac{1}{sin x}$ $cscx = 1/sinx$ )

$= 2 csc x$ $=2cscx$

Answer 2 · 2016-03-01T23:40:53

$\frac{2}{sin x}$ $2/sinx$

Explanation:

Write with a common denominator

$\frac{{sin}^{2} x + {(1 - cos x)}^{2}}{sin x (1 - cos x)}$ $(sin^2x + (1 - cosx)^2)/(sinx(1 - cosx))$

$= \frac{{sin}^{2} x + 1 - 2 cos x + {cos}^{2} x}{sin x (1 - cos x)}$ $=( sin^2x + 1 - 2cosx + cos^2x)/(sinx(1- cosx))$

$= \frac{{sin}^{2} x + {cos}^{2} x + 1 - 2 cos x}{sin x (1 - cos x)}$ $=( sin^2x + cos^2x + 1 - 2cosx)/(sinx(1-cosx))$

[using the identity : ${sin}^{2} x + {cos}^{2} x = 1]$ $sin^2x + cos^2x = 1]$

then becomes : $\frac{(1 + 1 - 2 cos x)}{sin x (1 - cos x)}$ $( (1 + 1 - 2cosx))/(sinx(1-cosx))$

$= \frac{2 (1 - cos x)}{sin x (1 - cos x)}$ $= (2(1 - cosx))/(sinx(1-cosx))$

$=( 2cancel(1-cosx))/(sinxcancel(1-cosx)) = 2/sinx$

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How do you simplify $(\frac{sin x}{1 - cos x}) + (\frac{1 - cos x}{sin x})$ $(sinx/(1-cosx))+((1-cosx)/sinx)$ ?

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