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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2Divide numerator and denominator by $\cos A \cos B$. – 2012-07-25
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0Oh I see now dividing by cos I get the correct answer thanks to all who posted. – 2012-07-25
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3One 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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0@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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0@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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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:
$$\dfrac{\dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y + \sin x\sin y}{\cos x\cos y}}$$
$$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y} +\dfrac{\cos x\sin y}{\cos x\cos y}} {\dfrac{\cos x\cos y} {\cos x\cos y}+ \dfrac{\sin x\sin y}{\cos x\cos y}}$$ $$\dfrac{\tan x+\tan y}{1+\tan x\tan y}$$