Question #0b5d8

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Question #0b5d8

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Answer 1 · 2017-12-05T11:48:19

Proof below

Explanation:

First know these three identities:
$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$
$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x=cos^2x-sin^2x$
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

First, we change the $sin 2 x$ $sin2x$ and $cos 2 x$ $cos2x$
$\frac{1 - sin 2 x}{cos 2 x} = \frac{1 - 2 sin x cos x}{{cos}^{2} x - {sin}^{2} x}$ $(1-sin2x)/(cos2x)=(1-2sinxcosx)/(cos^2x-sin^2x)$

Now this part may be hard to see, but you need to split the $1$ $1$ in the numerator using the third identity
$= \frac{{cos}^{2} x + {sin}^{2} x - 2 sin x cos x}{{cos}^{2} x - {sin}^{2} x}$ $=(cos^2x+sin^2x-2sinxcosx)/(cos^2x-sin^2x)$

Rearrange to look more like the expansion of a square
$= \frac{{cos}^{2} x - 2 sin x cos x + {sin}^{2} x}{{cos}^{2} x - {sin}^{2} x}$ $=(cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)$

Factorise the numerator, while using the difference two swaures the factor the denominator
$= \frac{{(cos x - sin x)}^{2}}{(cos x - sin x) (cos x + sin x)}$ $=(cosx-sinx)^2/((cosx-sinx)(cosx+sinx))$

Cancel like terms:
$= \frac{cos x - sin x}{cos x + sin x}$ $=(cosx-sinx)/(cosx+sinx)$

And we're done

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Question #0b5d8

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