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

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

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Answer 1 · 2015-05-19T02:05:17

Use the following facts:
$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$
$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$
$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x=cos^2x-sin^2x$
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

So you get:

$\frac{1 - \frac{sin x}{cos x}}{1 + \frac{sin x}{cos x}} = \frac{1 - 2 sin x cos x}{{cos}^{2} x - {sin}^{2} x}$ $(1-sinx/cosx)/(1+sinx/cosx)=(1-2sinxcosx)/(cos^2x-sin^2x)$
$(cosx-sinx)/cancel(cosx)*cancel((cosx))/cancel((cosx+sinx))=(1-2sinxcosx)/((cosx-sinx)cancel((cosx+sinx)))$

Taking: $(cosx-sinx)$ to the left side and using the fact that: $sin^2x+cos^2x=1$

$(cosx-sinx)^2=sin^2x+cos^2x-2sinxcosx$ which are indeed equals.

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

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