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

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

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Answer 1 · 2016-03-24T14:56:39

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Explanation:

Answer 2 · 2016-03-24T15:11:37

$cos 2 x = \frac{{cos}^{2} x - {sin}^{2} x}{{cos}^{2} x + {sin}^{2} x} = \frac{\frac{{cos}^{2} x}{{cos}^{2} x} - \frac{{sin}^{2} x}{{cos}^{2} x}}{\frac{{cos}^{2} x}{{cos}^{2} x} + \frac{{sin}^{2} x}{{cos}^{2} x}} = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x}$ $cos2x=(cos^2x-sin^2x)/(cos^2x+sin^2x)=(cos^2x/cos^2x-sin^2x/cos^2x)/(cos^2x/cos^2x+sin^2x/cos^2x)=(1-tan^2x)/(1+tan^2x)$

Answer 3 · 2016-03-24T15:16:48

Noah G

Mar 24, 2016

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

3 Answers

Explanation:

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How do you prove cos2x=1−tan2x1+tan2xcos 2x = (1-tan^2x)/(1+tan^2x)?

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