Prove that $\sqrt{\frac{1 - cos x}{1 + cos x}} \equiv \frac{1 - cos x}{| sin x |}$ $sqrt((1-cosx)/(1+cosx)) -= (1-cosx)/(|sinx|)$ ?

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Prove that $\sqrt{\frac{1 - cos x}{1 + cos x}} \equiv \frac{1 - cos x}{| sin x |}$ $sqrt((1-cosx)/(1+cosx)) -= (1-cosx)/(|sinx|)$ ?

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Answer 1 · 2018-02-14T00:27:27

Please see below.

Explanation:

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$\sqrt{\frac{1 - cos x}{1 + cos x}} = \sqrt{\frac{(1 - cos x) (1 - cos x)}{(1 + cos x) (1 - cos x)}} =$ $sqrt((1-cosx)/(1+cosx))=sqrt(((1-cosx)(1-cosx))/((1+cosx)(1-cosx)))=$

$\sqrt{\frac{{(1 - cos x)}^{2}}{1 - {cos}^{2} x}} = \sqrt{\frac{{(1 - cos x)}^{2}}{{sin}^{2} x}} = \frac{1 - cos x}{| sin x |}$ $sqrt(((1-cosx)^2)/(1-cos^2x))=sqrt((1-cosx)^2/sin^2x)=(1-cosx)/abssinx$

Answer 2 · 2018-02-14T00:38:55

We seek to prove that:

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

Consider the RHS:

$R H S = \frac{1 - cos x}{| sin x |}$ $RHS = (1-cosx)/(|sinx|)$

$\ \ \ \ \ \ \ \ = sqrt( ((1-cosx)/(|sinx|))^2 )$

$\ \ \ \ \ \ \ \ = sqrt( (1-cosx)^2/(sin^2x) )$

$\ \ \ \ \ \ \ \ = sqrt( (1-cosx)^2/(1-cos^2x) )$

$\ \ \ \ \ \ \ \ = sqrt( (1-cosx)^2/((1+cosx)(1-cosx) )$

$\ \ \ \ \ \ \ \ = sqrt( (1-cosx)/(1+cosx) )$

$\ \ \ \ \ \ \ \ = LHS \ \ \ \$ QED

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Prove that $\sqrt{\frac{1 - cos x}{1 + cos x}} \equiv \frac{1 - cos x}{| sin x |}$ $sqrt((1-cosx)/(1+cosx)) -= (1-cosx)/(|sinx|)$ ?

2 Answers

Explanation:

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2 Answers

Explanation:

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Prove that $\sqrt{\frac{1 - cos x}{1 + cos x}} \equiv \frac{1 - cos x}{| sin x |}$ $sqrt((1-cosx)/(1+cosx)) -= (1-cosx)/(|sinx|)$ ?