How do you verify $\frac{1 - cos x}{1 + cos x} = {(csc x - cot x)}^{2}$ $(1 - cos x) / (1 + cos x) = (csc x - cot x)^2$ ?

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How do you verify $\frac{1 - cos x}{1 + cos x} = {(csc x - cot x)}^{2}$ $(1 - cos x) / (1 + cos x) = (csc x - cot x)^2$ ?

1 Answer

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Answer 1 · 2015-11-04T14:41:51

Verify $\frac{1 - cos x}{1 + cos x} = {(csc x - cot x)}^{2}$ $(1 - cos x)/(1 + cos x) = (csc x - cot x)^2$

Explanation:

Multiply both numerator and denominator of left side by (1 - cos x).
$(\frac{1 - cos x}{1 + cos x}) (\frac{1 - cos x}{1 - cos x}) =$ $((1 - cos x)/(1 + cos x))((1 - cos x)/(1 - cos x)) =$
$= \frac{{(1 - cos x)}^{2}}{1 - {cos}^{2} x} = \frac{{(1 - cos x)}^{2}}{{sin}^{2} x} =$ $= (1 - cos x)^2/(1 - cos^2 x) = (1 - cos x)^2/(sin^2 x) =$
$= {(\frac{1 - cos x}{sin x})}^{2} = {(\frac{1}{sin x} - (\frac{cos x}{sin x}))}^{2} =$ $= ((1- cos x)/(sin x))^2 = (1/(sin x) - (cos x/sin x))^2 =$
$= {(csc x - cot x)}^{2}$ $= (csc x - cot x)^2$

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How do you verify $\frac{1 - cos x}{1 + cos x} = {(csc x - cot x)}^{2}$ $(1 - cos x) / (1 + cos x) = (csc x - cot x)^2$ ?

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Explanation:

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How do you verify $\frac{1 - cos x}{1 + cos x} = {(csc x - cot x)}^{2}$ $(1 - cos x) / (1 + cos x) = (csc x - cot x)^2$ ?