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    Hi,
    I'm stuck on a c3 trig question. Was wondering if any of you may be able to give some help.

    (Q) Prove the following identity:

    tan(a+45) + tan(a-45) = 2tan2a


    Many Thanks in advance,
    ayathullah
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    (Original post by ayathullah)
    Hi,
    I'm stuck on a c3 trig question. Was wondering if any of you may be able to give some help.

    (Q) Prove the following identity:

    tan(a+45) + tan(a-45) = 2tan2a


    Many Thanks in advance,
    ayathullah
    \tan (x+y) = \frac{\tan a + \tan y}{1-\tan x \tan y}.
    Use this to expand the LHS and then manipulate it until you have the RHS.
    Note that \tan (2a) = tan(a+a).
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    Hi,

    Just a little bit confused...........


    (Original post by JamieJ_93)

    = [- (tanA-1)^2 +(tanA+1)^2 ]/(1-tan^2A)
    Any chance you could tell me how you multiplied out the equation above.


    (Original post by JamieJ_93)

    = (2tanA + 2tanA ) / (1-tan^2A)
    And how did you get rid of the 1-tan^2A from the bottom to get to 2tan(A+A)

    Thanks,
    ayathullah
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    right, so first i can see already got , so using the identities for tan(a-b) and tan(a+b) with this..

    you end up with:

    [Tan(a) +1 ] / [1-Tan(a)] - [Tan(a) +1] / [1+ Tan(a)]

    then you combine this into one fraction to give

    [Tan(a) +Tan(a) +1+ [Tan(a)]squared ) + (Tan(a) +Tan(a) -1 - [Tan(a)]squared]

    all over: [1-[Tan(a)]squared -Tan(a) +Tan(a)]

    simplifying this you get:

    [4Tan(a)] / [1-[Tan(a)]squared]

    =2[2Tan(a)] / [1-[Tan(a)]squared]

    ( using the double angle identity: Tan(2a) = [2Tan(a)]/(1-[Tan(a)]squared)

    =2Tan(2a)

    hope you understood that, i have yet to master tex
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    i understand it now many thnaks for all your help
 
 
 
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