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Math Help - proving trigo identities

  1. #1
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    proving trigo identities

    1. Given that A=B+C, prove that tanA-tanB-tanC=tanAtanBtanC

    I can do it but if it's tan A + tan B...not minus.

    and I have a real problem with proving. problem is, I don't know where to start from if I'm trying to solve a 'prove' question. any advice?

    thank you so much!
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  2. #2
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    Quote Originally Posted by colloquial View Post
    1. Given that A=B+C, prove that tanA-tanB-tanC=tanAtanBtanC

    I can do it but if it's tan A + tan B...not minus.

    and I have a real problem with proving. problem is, I don't know where to start from if I'm trying to solve a 'prove' question. any advice?

    thank you so much!
    It's just a rewrite of the tan of a sum identity:

    <br />
\tan(B+C) = \frac{\tan(B)+\tan(C)}{1-\tan(B)\tan(C)}<br />

    so:

    <br />
\tan(A) = \frac{\tan(B)+\tan(C)}{1-\tan(B)\tan(C)}<br />

    rearranging:

    <br />
\tan(A) - \tan(A) \tan(B) \tan(C)= \tan(B)+\tan(C)<br />

    then:

    <br />
\tan(A) - \tan(B) -\tan(C)=\tan(A)\tan(B)\tan(C)<br />

    Here we start from the known, and work towards what we have to prove.

    RonL
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  3. #3
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    Hello, colloquial!

    This is easier than you think . . .


    1. Given that A\:=\:B+C, prove that: . \tan A -\tan B - \tan C \:=\:\tan A\tan B\tan C

    We have been given: . A\;=\;B+C

    Take the tangent of both sides: . \tan A \;=\;\tan(B + C)

    . . and we have: . \tan A \;=\;\frac{\tan B + \tan C}{1 - \tan B\tan C}\quad\Rightarrow\quad \tan A \,-\, \tan A\tan B\tan C \;=\;\tan B \,+\,\tan C

    Therefore: . \tan A - \tan B - \tan C \;=\;\tan A\tan B\tan C


    Edit: Too fast for me, Cap'n!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    Edit: Too fast for me, Cap'n!
    Hope this helps:

    "I woke up this mornin'
    with those beaten to MathHelpForum reply blues <der dar dur dum>..."

    CFB
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    CFB
    is that an abbreviation of your blues name? what's it stand for?
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  6. #6
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    thank you!
    oh dear lord I never expected it to be this easy, I went all the way:
    expanding tanAtanBtanC to thinking that A+B+C=180 (although it wasn't stated).

    proving: not my favourite.
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  7. #7
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    Hello, colloquial!

    . . . thinking that A+B+C\,=\,180 (although it wasn't stated).
    The other identity you referred to is one of my favorites.
    . . I also started with angles A,\,B,\,C in a triangle . . . *blush*

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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    is that an abbreviation of your blues name? what's it stand for?
    See the Blues thread.

    RonL
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