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Proving two sides of equation for triangles

  1. Nov 6, 2007 #1
    [SOLVED] Proving two sides of equation for triangles

    1. The problem statement, all variables and given/known data
    In angle ABC, which is an isosceles triangle with <B = <C, show that
    2cot(a) = tan(b) = cot(b)


    2. Relevant equations
    tan2a = 2tana / 1 - tan^2 a

    tan (x - y) = tanx - tany / 1 + tanx tany


    3. The attempt at a solution

    Since it is isosceles, that means two sides are equal, therefore, <A = 180 - 2B

    2cot(A) = 2cot (180 - 2B)
    = 1 / 2tan(180 - 2B)
    = 1 / 2(tan180 - tan2B / 1 + tan180 tan2B)
    = 1 / 2(-tan2B)
    = 1 / -2tan2B)
    = 1 / -2(2tanB / 1 - tan^2 B)

    After this, I get confused. Can someone please tell me if I am doing this right? Please help. Thanks.
     
  2. jcsd
  3. Nov 6, 2007 #2

    learningphysics

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    that's your mistake.

    2*cot(180 - 2B) = 2*[1/tan(180-2B)] = 2/tan(180-2B)

    also, please uses parentheses... in your work here especially when dealing with fractions and sums in the numerator and denominator. people will get confused as to your exact meaning: for example this:

    means: tan180 - tan2B + (tan180)*(tan2B) the way you wrote it.

    you should write:

    (tan180 - tan2B) / (1 + tan180 tan2B )
     
  4. Nov 6, 2007 #3
    Thanks for your help. Here is what I am getting:

    2cot(a) = 2/tan(180-2B)
    = 2 / {(tan180 - tan2b) / (1 + tan180tan2b)}
    = 2 / {-tan2b / 1}
    = 2 / {-2tanb / 1 - tan^2 b)

    After this, i am still confused as to how to get the equation to equal to tanb - cotb.

    Please do help. Thanks.
     
  5. Nov 6, 2007 #4

    learningphysics

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    2 / [-2tanb / (1 - tan^2 b)]

    = [tex]2*(\frac{1-tan^2(b)}{-2tan(b)})[/tex]

    =[tex]\frac{1-tan^2(b)}{-tan(b)}[/tex]

    = -1/tan(b) + tan(b)
    = -cot(b) + tan(b)
    = tan(b) - cot(b)
     
  6. Nov 6, 2007 #5
    Wow, thanks very much learningphysics. You have helped me so much and even shown me the answer. I was worried about the negative sign and how to get rid of it, but as soon as I saw your solution, I understood quickly the idea behind it. Thanks very much again.
     
  7. Nov 6, 2007 #6

    learningphysics

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    no prob. you're welcome.
     
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