Question #6a1a5

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Question #6a1a5

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Answer 1 · 2017-03-29T23:35:14

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

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

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

$= \frac{2 sin x cos x}{3 {sin}^{2} x - {cos}^{2} x}$ $=(2sinxcosx)/(3sin^2x-cos^2x)$

$= \frac{\frac{2 sin x cos x}{{cos}^{2} x}}{\frac{3 {sin}^{2} x}{{cos}^{2} x} - \frac{{cos}^{2} x}{{cos}^{2} x}}$ $=((2sinxcosx)/cos^2x)/((3sin^2x)/cos^2x-cos^2x/cos^2x)$

$= \frac{2 tan x}{3 {tan}^{2} x - 1}$ $=(2tanx)/(3tan^2x-1)$

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