1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted