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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