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

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

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Answer 1 · 2016-03-20T21:39:02

see explanation

Explanation:

Manipulating the left side using $Double angle formulae$ $color(blue)" Double angle formulae "$

$∙ sin 2 x = 2 sin x cos x$ $• sin2x = 2sinxcosx$

$∙ cos 2 x = {cos}^{2} x - {sin}^{2} x$ $• cos2x = cos^2x - sin^2x$

and using ${sin}^{2} x + {cos}^{2} x = 1 we can also obtain$ $sin^2x + cos^2x = 1 " we can also obtain "$

$cos 2 x = (1 - {sin}^{2} x) - {sin}^{2} x = 1 - 2 {sin}^{2} x$ $cos2x = (1 - sin^2x) - sin^2x = 1 - 2sin^2x$

and $cos 2 x = {cos}^{2} x - (1 - {cos}^{2} x) = 2 {cos}^{2} x - 1$ $cos2x = cos^2x - (1 - cos^2x ) = 2cos^2x - 1$

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

$= (cancel(2) sinx cancel(cosx))/(cancel(2) cancel(cosx) cosx)= (sinx)/(cosx) = tanx = " right side "$

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

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How do you prove (sin 2x) / (1 + cos2x) = tan xsin2x1+cos2x=tanx(sin 2x) / (1 + cos2x) = tan x?

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