Proving identities

**hachm361** · November 12th 2012, 05:38 PM

2tanA / 1+tan²A = sin2A

I know that 2tanA is tan a + tan a and that sin2A is 2sinAcosA

Could someone help me figure out the rest?

**MarkFL** · November 12th 2012, 06:37 PM

Traditionally, when proving identities you want to begin with one side (usually the left) and apply known standard identities and perhaps some algebra so that the right side results. Looking at the left side, I suggest you begin with the Pythagorean identity $1+\tan^2\theta=\sec^2\theta$ .

**darthjavier** · November 12th 2012, 06:49 PM

Hi

$\\\mathrm{\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}}\\\mathrm{\tan(2A)=\dfrac{2\tan A}{1-\tan^2A}}\\\mathrm{\dfrac{\sin2A}{\cos2A}=\dfrac{2 \tan A}{1-\tan^2A}}\\\mathrm{\sin2A=\cos2A\times\dfrac{2\tan A}{1-\tan^2A}}\\\mathrm{\sin2A=\dfrac{\cos^2A-\sin^2A}{\cos^2A+\sin^2A}\times\dfrac{2\tan A}{1-\tan^2A}}\\\mathrm{\sin2A=\dfrac{1-\tan^2A}{1+\tan^2A}\times\dfrac{2\tan A}{1-\tan^2A}}\\\mathrm{\sin2A=\dfrac{2\tan A}{1+\tan^2 A}}$

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