# Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$

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 $\cos A \cos B$. – Aang Jul 25 '12 at 17:26
Oh I see now dividing by cos I get the correct answer thanks to all who posted. – El Bananero Jul 25 '12 at 17:31
One can prove and identity or solve an equation. But to speak of solving an identity could leave some doubt about what you mean. – Michael Hardy Jul 25 '12 at 18:15
@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 $A,B$ – Ross Millikan Jul 25 '12 at 18:20
@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. – Rick Decker Jul 25 '12 at 19:56

Now divide by $\cos A \cos B$ and you are there
$$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin x\sin y}{\cos x\cos y}}$$ $$\dfrac{\tan x+\tan y}{1+\tan x\tan y}$$