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Math Help - Urgent Help With Trig Proofs

  1. #1
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    Urgent Help With Trig Proofs

    (Sinθ - 1)/(1-sin2θ) = (cscθ)/(-cscθ-1)

    or

    (1-sinθ)/(1+sinθ) = (tanθ-secθ)^2
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  2. #2
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    1) Please write what you mean. You have written \sin(2\theta). I am guessing this is not your intent.

    2) Are you acquainted with the Pythagorean Identity? \cos^{2}(x) + \sin^{2}(x) = 1?

    3) \csc(\theta) = 1/\sin(\theta)
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  3. #3
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    i intended it to read sin^2 theta
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by rabidsquirrel View Post
    (Sinθ - 1)/(1-sin2θ) = (cscθ)/(-cscθ-1)
    \frac{\csc x}{-\csc x - 1} = - \frac{1}{1+\sin x}

    you should know what to multiply then...
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  5. #5
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    Quote Originally Posted by kalagota View Post
    \frac{\csc x}{-\csc x - 1} = - \frac{1}{1+\sin x}

    you should know what to multiply then...
    you cannot multiply across, it is a proof
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  6. #6
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    Quote Originally Posted by kalagota View Post
    \frac{\csc x}{-\csc x - 1} = - \frac{1}{1+\sin x}

    you should know what to multiply then...
    sorry misread your reply
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by rabidsquirrel View Post
    sorry misread your reply
    no problem.. are you done?
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  8. #8
    Senior Member nikhil's Avatar
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    Lightbulb Here it is

    T=[sin()-1]/[1-sin()^2]=[sin()-1]/[(1-sin())(1+sin()] as [a^2-b^2=(a+b)(a-b)]
    =-1/1+sin() now dividing both numerator and denominator by sin() we get
    T=-csc()/1+csc()=csc()/-1-csc(). Hence proved.
    2))A=(1-sin())/(1+sin()) multiplying both numerator and denominator by 1-cos() we get A=(1-sin())^2/cos()^2 or [(1-sin())/cos()]^2 which is=[sec()-tan()]^2 which is= [tan()-sec()]^2 hence proved.
    Note (a-b)^2=(b-a)^2
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