  <p>Ha ha! Fools! I have lured you into none other than a homework help topic! And this is no ordinary homework help topic. No, your test is a mere pre-calculus trigonometry verification that this student can't seem to grasp. And now you must solve it for me. And don't even try to leave. It's not like you can just click the back button or anything.</p>

<p>(cosA+sinA)/(cosA-sinA)= (1+2sinA)/(cos2A)</p>

<p>Uh... any ideas or clues or anything? Even that would be greatly appreciated. Thanks.</p>

<p>Verifications are so much fun you should try solving it yourself.

<p>I was thinking of multiplying and dividing LHS by (Cos A + Sin A), but you'd get 1+Sin2A in the numerator instead of 1+2SinA, I think. Not sure, I'd have to work it out on paper.</p>

<p>first multiply the left by the conjugate cosA+sinA then go</p>

<p>Wow, so close, I got (1+2sinAcosA)/cos2A</p>

<p>Truffliepuff that's what i did but it didn't work :(</p>

<p>Oh great SuperModerators, smite this sinner!</p>

<p>Oh wait nvm I forgot sin2A=2sinAcosA lol</p>

<p>yes it does.</p>

<p>(cosA+sinA)/(cosA-sinA)= (1+2sinA)/(cos2A)
x cosA+sinA</p>

<p>cos^2A+2cosasinA+sin^2A / cos^2A-sin^2A = 1+2sinA / cos2A</p>

<p>(sin^2A+cos^2A = 1)</p>

<p>1+2cosAsinA / 2 cosA = 1+2sinA / 2cosA</p>

<p>TOP Part is equal because 2 sin A = 2cosAsinA
BOTTOM part is equal because 2cosA=2cosA literally</p>

<p>^
Sin 2A = 2 Sin A Cos A
Cos 2A = Cos^2 A - Sin^2 A</p>