# Trigonometry

Prove that
1-cosA+cosB-cos(A+B)/1+cosA-cosB-cos(A+B) = tanA/2cotA/2

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1. you've posted a lot of these lately. What have you tried so far on this one?

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2. hmm. I see most of them have not been answered yet. So, what the heck? It's a slow day.

First off, a typo. You should have said

... = tanA/2 cotB/2

I gotta admit, I had a hard time with this one, just manipulating the LS. However,

tanA/2 = sinA/(1+cosA)
cotB/2 = sinB/(1-cosB)
so,
tanA/2 cotB/2 = sinA sinB/(1 + cosA - cosB - cosAcosB)

also,

tanA/2 = (1-cosA)/sinA
cotB/2 = (1+cosB)/sinB
so,
tanA/2 cotB/2 = (1 - cosA + cosB - cosAcosB)/sinAsinB

that gives us

tan^2 A/2 * cot^2 B/2 =

1 - cosA + cosB - cosA cosB
---------------------------
1 + cosA - cosB - cosA cosB

Now, that means that we have to show that

(1-cosA+cosB-cos(A+B))^2 = 1 - cosA + cosB - cosA cosB
and
(1+cosA-cosB-cos(A+B))^2 = 1 + cosA - cosB - cosA cosB

If you expand that out and collect terms, you will see that it is true, but it's a lot of work.

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3. using the half-angle formulas, you can show that

1 - cosA + cosB - cosA cosB
= 2[sin^2 A/2 + cos^2 B/2 - cos^2 (A+B)/2]

Then, with a little more fiddling, that can be changed to

sin A/2 cos B/2 sin(A+B)/2

and the denominator comes out to

cos A?2 sin B/2 sin(A+B)/2

divide those two expressions and you get

sin A/2 cos B/2
---------------------
cos A/2 sin B/2

= tanA/2 cotB/2

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