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.
Prove $\frac{\sin(A+B)}{\cos(A-B)}=\frac{\tan A+\tan B}{1+\tan A\tan B}$
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trigonometry
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0@RickDecker: No big problem. I put it back. – 2012-07-25
2 Answers
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Now divide by $\cos A \cos B$ and you are there
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What i get is, how to solve the problem?? Is that correct then here u are:
$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$
$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin x\sin y}{\cos x\cos y}}$ $\dfrac{\tan x+\tan y}{1+\tan x\tan y}$