Can someone help me to prove this? (cosA+cosB)/(cosA-cosB)=cot((1/2)(A-B))cot((1/2)(A+B))

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Can someone help me to prove this? (cosA+cosB)/(cosA-cosB)=cot((1/2)(A-B))cot((1/2)(A+B))

Mathematics
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we can use a few identities to rewrite the right hand side. First right it as product of two cosines over product of two sines. Then I used cos(u +/- v)= cos(u)cos(v) -/+ sin(u)sin(v) sin(u +/- v)= sin(u)cos(v) +/- cos(u)sin(v) sin^2(x)=(1-cos(2x))/2 cos^2(x)=(1+cos(2x))/2. its just a matter of running the algebra then. Let me know if you need help with any steps or getting started.
write*
yeah I need help. Im really confused with this

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Please can you show me the work?
Trying to learn trigonometry
ok the right hand side can be written... {cos((A-B)/2)*cos((A+B)/2)}/{sin((A-B)/2)*sin((A+B)/2)}
each term is in the form of the first 2 identities that i posted
so if you write them out and then multiply you'll get terms in the form of the 3rd and 4th identities I posted
It all simplifies down to the left hand side
let me check it out. I will write you if i get lost. Thanks
Yep let me know.
http://www.sosmath.com/trig/Trig5/trig5/trig5.html here are a bunch of identities
These trig problems are always mindless manipulation of symbols. I'll have a go at it and get back to you shortly.
got it!
good :)
OK. The sum and difference formulae for cosines are as follows: CosA + CosB = 2cos((A+B)/2)cos((A-B)/2) and CosA - CosB = -2sin((A+B)/2)sin((a-B)/2) So, your LHS = -cot((A+B)/2)cot((A-B)/2) Note. I have a minus sign in front of your solution.
i had a negative as well
yeah the solution has a - sign
emun i answered one of your old physics questions, when you were asking about dimensions you meant units right?
yeah. I dint know how to put that question into english. And your right, i ment units.
ok, I've heard units called dimensions before I was just making sure that's what you meant.
In spanish we say dimension. I supposed it was the same in English. lol

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