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    Struggling with trig again can anyone help me
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    Name:  image.jpg
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Size:  21.5 KBThis is the transition I don't understand
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    first step you flip both sides of the equation then in the next step you mutiply the top and bottom of the fraction by (secθ -tanθ)
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    (Original post by Alex621)
    first step you flip both sides of the equation then in the next step you mutiply the top and bottom of the fraction by (secθ -tanθ)
    Why would you multiply by secθ -tanθ?


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    (Original post by Abby5001)
    Why would you multiply by secθ -tanθ?


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    to perform the transition...

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    (Original post by Abby5001)
    Why would you multiply by secθ -tanθ?


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    If you then expand the brackets you'll be able to use a trig identity to simplify the fraction
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    (Original post by Alex621)
    If you then expand the brackets you'll be able to use a trig identity to simplify the fraction
    Do you mind explaining in a bit more detail please step by step?


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    (Original post by Abby5001)
    Do you mind explaining in a bit more detail please step by step?


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    Once you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)


    Expand (secθ + tanθ) (secθ - tanθ), to get sec2θ - tan2θ.

    Now use the identity sec2θ = tan2θ + 1, to simplify it
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    (Original post by Alex621)
    Once you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)


    Expand (secθ + tanθ) (secθ - tanθ), to get sec2θ - tan2θ.

    Now use the identity sec2θ = tan2θ + 1, to simplify it
    Sorry I wasn't clear, I just meant can you explain how you get to 1/x = (secθ -tanθ) / (secθ + tanθ) (secθ - tanθ)


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    (Original post by Abby5001)
    Do you mind explaining in a bit more detail please step by step?


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    You are trying to get the difference of two squares on the denominator.

    It's a bit like rationalising the denominator when you multiply by the conjugate.
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    (Original post by Muttley79)
    You are trying to get the difference of two squares on the denominator.

    It's a bit like rationalising the denominator when you multiply by the conjugate.
    Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you


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    (Original post by Abby5001)
    Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you


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    I used to hate trig questions in c3 because I never knew where to start but once you've looked at lots of questions you start to get an idea of what you need to do
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    (Original post by Alex621)
    I used to hate trig questions in c3 because I never knew where to start but once you've looked at lots of questions you start to get an idea of what you need to do
    That's good to know, so is multiplying by the conjugate something that is prevalent in c3 because I have never seen it aside from in surds?


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    (Original post by Abby5001)
    That's good to know, so is multiplying by the conjugate something that is prevalent in c3 because I have never seen it aside from in surds?


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    I've never seen it before apart from surds but maybe different exam boards use it with trig?
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    (Original post by Abby5001)
    Right I think I see it's just that I would never think to do this in an exam so it's quite confusing but thank-you


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    I think I've done this question recently - the method the markscheme used seem a little convoluted to me, I much prefer what I did!

    x = \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta} and the continue from there converting everything into sines and cosines, collecting and then seperating and converting back to secs. (x + 1/x = 2 sec theta, right?)
 
 
 
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