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How would I prove the following trig identity?

$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$

My work thus far has been: $$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$

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    $\frac{cosA+cosB}{cosA-coSB}=\frac{2cos\frac{A-B}{2}cos\frac{A+B}{2}}{-2sin\frac{A-B}{2}sin\frac{A+B}{2}}=-cot\frac{A-B}{2}cot\frac{A+B}{2}$, dividing the numerator & the denominator by $sin\frac{A-B}{2}sin\frac{A+B}{2}$, right?2012-08-04
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    I'm probably being stupid here, but if this were true, wouldn't [this](http://www.wolframalpha.com/input/?i=%5Bcos%280.6%2B0.4%29%2F%28cos%280.6%29-cos%280.4%29%29%5D+%2B+%5Bcot%280.5%29%280.2%29cot%280.5%29%281%29%5D) be $0$.2012-08-04
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    @lab bhattacharjee how would I divide by sin(a-b)/22012-08-04
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    Just a minor nit-pick: you **solve equations**, rather than **prove** them. Trigonometric things that you prove are identities rather than equations.2012-08-04
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    Sorry I meant identity......2012-08-04
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    @Old: It's not such a minor nitpick in my view. Misuse of "prove" is one of the major English errors on this site, and I think a certain amount of insistence on correct use of terminology will be required if we don't want the bad examples to become so abundant that people start learning wrong mathematical English from this site.2012-08-04
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    I agree fully! I was just being polite, being something of a a newbie here.2012-08-04
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    Use the cotangent half-angle formula or the tangent half-angle formula.2012-08-04

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