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Math Help - [SOLVED] Prove: tan(A+45) = (1+tanA)/(1-tanA)

  1. #1
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    [SOLVED] Prove: tan(A+45) = (1+tanA)/(1-tanA)

    Prove: tan(A+45) = (1+tanA)/(1-tanA)

    Please i need help, i know that you start with the RHS ... but i do not know.

    Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.

    Thanks so much
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    Prove: tan(A+45) = (1+tanA)/(1-tanA)

    Please i need help, i know that you start with the RHS ... but i do not know.

    Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.

    Thanks so much
    Use the sum of angles formula:
    tan(A + 45) = \frac{tan(A) + tan(45)}{1 - tan(A)~tan(45)}

    = \frac{1 + tan(A)}{1 - tan(A)}

    -Dan
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    I do not understand that ...

    i need it to be :

    RHS = (1+tanA)/(1-tanA)

    =(1+sinA/cosA)/(1-sinA/cosA) etc.

    i am not saying the above is wrong, i just need it like this, coz i have no clue wot u said lol
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    I do not understand that ...

    i need it to be :

    RHS = (1+tanA)/(1-tanA)

    =(1+sinA/cosA)/(1-sinA/cosA) etc.

    i am not saying the above is wrong, i just need it like this, coz i have no clue wot u said lol
    The only other way I can think to help you with this is to use the sum of angles formula for sine and cosine:
    sin(A + B) = sin(A)~cos(B) + sin(B)~cos(A)
    and
    cos(A + B) = cos(A)~cos(B) - sin(A)~sin(B)

    Then
    tan(A + 45) = \frac{sin(A + 45)}{cos(A + 45)}

    = \frac{sin(A)~cos(45) + sin(45)~cos(A)}{cos(A)~cos(45) - sin(A)~sin(45)}
    which can be reduced to the needed expression.

    I don't think that there is any way to do this problem without using one of these methods.

    -Dan
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    i am stuck, i actually thought i had it ... please help again, soz
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  6. #6
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    Quote Originally Posted by topsquark View Post
    sin(A + B) = sin(A)~cos(B) + sin(B)~cos(A)
    and
    cos(A + B) = cos(A)~cos(B) - sin(A)~sin(B)

    Then
    tan(A + 45) = \frac{sin(A + 45)}{cos(A + 45)}

    = \frac{sin(A)~cos(45) + sin(45)~cos(A)}{cos(A)~cos(45) - sin(A)~sin(45)}
    The only thing I can think of to do is continue:
    = \frac{\frac{\sqrt{2}}{2}~sin(A) + \frac{\sqrt{2}}{2}~cos(A)}{\frac{\sqrt{2}}{2}~cos(  A) - \frac{\sqrt{2}}{2}~sin(A)}

    = \frac{sin(A) + cos(A)}{cos(A) - sin(A)}

    Now divide numerator and denominator by cos(A).

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Now divide numerator and denominator by cos(A).
    Shouldnt we consider the case cos(A) = 0 separately also? I was wondering whether we could do this case without invoking inverse trig functions. I didnt get any idea
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    please can you do the entire sum, like just add the rest on as i have no clue what is going on,
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Shouldnt we consider the case cos(A) = 0 separately also? I was wondering whether we could do this case without invoking inverse trig functions. I didnt get any idea
    Since cos(0) = 1 there is no problem with any undefined expressions. Why do you bring that up as a specific case?

    -Dan
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jono Mathematician View Post
    please can you do the entire sum, like just add the rest on as i have no clue what is going on,
    The only thing left to do is divide the numerator and denominator by cos(A). If you can't do this all at once, then take the numerator and denominator separately and do the division. I presume you know that tan(A) = sin(A) / cos(A), right?

    -Dan
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  11. #11
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    I have spent the last 4 hours doing 15 or so trig "proving identities" and i really cannot get this one, even after u have said that, i am at the last stage with sina+cosA/cosa-sina...

    please can you just show me in numbers, i aint good with reading numbers, as in divide numerator by x etc, i need to see it to understand ...thanks
    [/font]
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  12. #12
    Xei
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    You don't need to use sin or cos anyway, the tan(A+B) formula constitutes a sufficient proof, and looking at the question, that is probably the one that you are supposed to use.

    Start on the left and work to the right. Either use the sin and cosine addition formulae

    sin(A+B)=sinAcosB+sinBcosA
    cos(A+B)=cosAcosB-sinAsinB

    and the identity tanx=sinx/cosx

    which is the same as saying tan(A+B)=sin(A+B)/cos(A+B)

    then the rest should be quite easy, just replace the sin and cos in the fraction with the identities above.

    Alternatively and much easier, there is an additional identity,

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

    which you can take to be true (it's proved easily by expressing tan as sin/cos but you don't need to do this).
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