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Math Help - trig proof

  1. #1
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    trig proof

    1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

    Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by cute_poison View Post
    1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

    Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!

    Ok, before I work on this, let me make sure I am working the problem you have because this:

    \frac{1 + \sin{x}}{1 - \sin{x}}  = \left(\frac{1 + \tan{\frac{x}{2}}}{1 + \tan{\frac{x}{2}}}\right)^{2}

    is not necessarily true, since

    \frac{1 + \sin{x}}{1 - \sin{x}}  = 1

    is not always true.
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by sinewave85 View Post
    Ok, before I work on this, let me make sure I am working the problem you have because this:

    \frac{1 + \sin{x}}{1 - \sin{x}}  = \left(\frac{1 + \tan{\frac{x}{2}}}{1 + \tan{\frac{x}{2}}}\right)^{2}

    is not necessarily true, since

    \frac{1 + \sin{x}}{1 - \sin{x}}  = 1

    is not always true.
    Your RHS simplifies to 1 so I'd imagine the OP meant

    \frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{1+tan^2({\frac{x}{2}})  }
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Your RHS simplifies to 1 so I'd imagine the OP meant

    \frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{1+tan^2({\frac{x}{2}})  }
    Yeah, I read it wrong, thanks.
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  5. #5
    Member sinewave85's Avatar
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    I hope what the op meant was:

    \frac{1 + \sin{x}}{1 - \sin{x}} = \frac{1+tan({\frac{x}{2}})}{[1+tan({\frac{x}{2}})]^{2}}<br />

    since that is how I think it reads.
    (The square is outside the brackets for the right-hand denominator).
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  6. #6
    Member sinewave85's Avatar
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    In either case, you are going to need the half-angle identities for tan, which are:

    \tan{\frac{x}{2}} = \frac{1 - \cos{x}}{\sin{x}}

    \tan{\frac{x}{2}} = \frac{\sin{x}}{1 + \cos{x}}
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  7. #7
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    Trig Identity

    Hello cute_poison
    Quote Originally Posted by cute_poison View Post
    1+sinx/1-sinx=((1+tan(x/2))/(1+tan(x/2))^2

    Need to prove it but cant get my head round it!!if anyone could help asap!!thanks!!
    I think you must have made a typo in your original posting, but perhaps this is what you meant:

    Prove the identity \frac{1+\sin x}{1-\sin x} =\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2

    If t = \tan \left(\frac{x}{2}\right) then \sin x = \frac{2t}{1+t^2}. (For instance, see Half Angle Formula)

    \Rightarrow \frac{1+\sin x}{1-\sin x}=\frac{1+\frac{2t}{1+t^2}}{1-\frac{2t}{1+t^2}}

    =\frac{(1+t^2)(1+\frac{2t}{1+t^2})}{(1+t^2)(1-\frac{2t}{1+t^2})}

     =\frac{(1+t^2)+2t}{(1+t^2)-2t}

    = \frac{1+2t+t^2}{1-2t+t^2}

    = \frac{(1+t)^2}{(1-t)^2}

    = \left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right)^2

    Does that help?

    Grandad
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