How to solve this identity? Thank you!

Question

$(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)) = (sin(2a))/(sin(2b))$

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Answer 1 · 2017-12-10T13:53:07

We know that, $tana+-tanb$

$=sina/cosa+-sinb/cosb$

$=(sinacosb+-cosasinb)/(cosacosb$

$=sin(a+-b)/(cosa*cosb)$

So, $(tana+tanb)/(tana-tanb)=sin(a+b)/sin(a-b)$

$LHS=(tan(a+b)+tan(a-b))/(tan(a+b)-tan(a-b)$

$=sin(a+b+a-b)/(sin(a+b-(a-b))$

$=sin(2a)/(sin(2b))=RHS$