How do you prove $\frac{1 - sin 2 x}{cos 2 x} = \frac{cos 2 x}{1 + sin 2 x}$ $(1 - sin2x) /(cos2x) = (cos2x) / (1 + sin2x)$ ?

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How do you prove $\frac{1 - sin 2 x}{cos 2 x} = \frac{cos 2 x}{1 + sin 2 x}$ $(1 - sin2x) /(cos2x) = (cos2x) / (1 + sin2x)$ ?

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Answer 1 · 2015-05-14T09:26:46

We have to prove that $\frac{1 - sin 2 x}{cos 2 x} = \frac{cos 2 x}{1 + sin 2 x}$ $(1-sin2x)/(cos2x)=(cos2x)/(1+sin2x)$

To do this we transform left side:

$\frac{1 - sin 2 x}{cos 2 x} = \frac{({sin}^{2} x + {cos}^{2} x) - 2 sin x cos x}{cos 2 x}$ $(1-sin2x)/(cos2x)=((sin^2x+cos^2x)-2sinxcosx)/(cos2x)$

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

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

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

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

Now we expand the expresion by multiplying both numerator and denominator by $(cos x + sin x)$ $(cosx+sinx)$

So we get:

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

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

$= \frac{cos 2 x}{1 + sin 2 x}$ $=(cos2x)/(1+sin2x)$

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How do you prove $\frac{1 - sin 2 x}{cos 2 x} = \frac{cos 2 x}{1 + sin 2 x}$ $(1 - sin2x) /(cos2x) = (cos2x) / (1 + sin2x)$ ?

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How do you prove $\frac{1 - sin 2 x}{cos 2 x} = \frac{cos 2 x}{1 + sin 2 x}$ $(1 - sin2x) /(cos2x) = (cos2x) / (1 + sin2x)$ ?