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Proving trig equations using addition formulae

  1. Sep 22, 2013 #1
    SOLVED! :)

    1. The problem statement, all variables and given/known data

    The question is as follows;

    Prove that (tan(A+B)-tanA)/1+tan(A+B)tanA = tanB


    2. Relevant equations

    I'm certain the addition fomulae need to be applied, although I'm not entirely sure how. I genuinely have tried many times!

    3. The attempt at a solution

    Okay, my first attempt was simply to work with the LHS, rewriting tan(A+B) as (tanA+tanB)/1-tanAtanB on both the top and bottom. I then tried to write tanA as sinA/cosA but this just made the LHS really messy and I wasn't able to cancel anything down.

    I then tried to multiply both sides by (1+tan(A+B)tanA), leaving me with (tan(A+B)-tanA) = tanB(1+tan(A+B)tanA). I then expanded the brackets, rewrote all the tans as sin/cos and then attempted to cancel, but this again proved useless as I just ended up with a really really really messy equation.

    I've tried this question many times and I still haven't managed to prove it. I am sure this is something very small that I'm missing/forgetting as this is the first trig question that I'm genuinely struggling with. I would appreciate if someone could nudge me in the right direction. Thank you in advance :).
     
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2

    BruceW

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    Homework Helper

    as you say, the key is the addition formula. So try getting tan(A+B) on one side of the equation, and see if the other side agrees with the addition formula.
     
  4. Sep 22, 2013 #3
    Okay, so I had a go at getting tan(A+B) on one side but I only ended up going in circles again. Perhaps it would help if I showed you how I got lost.

    First I moved 1+tan(A+B)tanA over to get
    tan(A+B) - tanA = tanB + tan(A+B)tanAtanB ... This is where I suspect I may be going wrong

    Then I rearragned to get
    tan(A+B) - tan(A+B)tanAtanB = tanA + tanB

    Okay, so this is the part where I start losing it and going in circles

    Factoring out tan(A+B) I get
    tan(A+B)(1-tanAtanB) = tan(A+B)

    ∴ tanAtanB = 2

    And I'm certain that's not correct. Perhaps this makes it easier for you to spot wher I've gone wrong. Thanks again


    EDIT: I've managed to crack it, I just forgot to implement the formula - silly me! Thank you so much for pointing me in the right direction though :).
     
    Last edited: Sep 22, 2013
  5. Sep 22, 2013 #4

    eumyang

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    Homework Helper

    First, you should have clarified in the beginning the original problem, because what you wrote:
    (tan(A+B)-tanA)/1+tan(A+B)tanA = tanB
    looks like this:
    [itex]\frac{\tan (A+B) - \tan A}{1} + \tan (A+B) \tan A = \tan B[/itex]

    It looks like you changed the right side from "tan A + tan B" to "tan (A+B)". Why? You can't do that. Leave it as it was; you're last step should be
    [itex]\tan (A+B)(1 - \tan A \tan B) = \tan A + \tan B[/itex]
    So what happens next?

    EDIT: OP edited his/her previous post while I was writing mine. :)
     
  6. Sep 22, 2013 #5
    Apologies for not making the question clear, I forgot to put some bracket in :eek:

    And even though you were giving instruction for solving a different problem to the one I thought I asked, you managed to help me!

    I moved across the (1+tan(A+B)tanA) and then brought the tan(A+B) to one side like you said

    I ended up with
    tan(A+B)-tan(A+B)tanAtanB = tanB+tanA

    Factoring out tan(A+B) I got
    tan(A+B)(1-tanAtanB) = tanA + tanB

    Rewriting tan(A+B) as tanA + tanB/(1-tanAtanB) I got
    tanA + tanB/(1-tanAtanB)*(1-tanAtanB) =tanA+tanB

    Cancelling the (1-tanAtanB)'s left me with
    tanA+tanB = tanA+tanB

    I think that is correct, as I've proved both sides to be equal.

    Again, thank you very much.
     
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