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How would I solve the following double angle identity. $$ \frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B} $$ So far my work has been. $$ \frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B+\sin A\sin B} $$ But what would I do to continue.

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    Divide numerator and denominator by $\cos A \cos B$.2012-07-25
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    Oh I see now dividing by cos I get the correct answer thanks to all who posted.2012-07-25
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    One can _prove_ and _identity_ or _solve_ an _equation_. But to speak of _solving_ an _identity_ could leave some doubt about what you mean.2012-07-25
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    @Rick Decker: please do not change the variables from $A,B$ to $x,y$ as avatar's comment and my answer were in terms of $A,B$2012-07-25
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    @Ross. Sorry about that. avatar's comment, your answer, and my edit came virtually on top of each other; I didn't see the notifications while I was editing. Note to self: for questions that are likely to be answered immediately after they're posted, delay editing until the dust settles.2012-07-25
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    @RickDecker: No big problem. I put it back.2012-07-25

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