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    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

    Need help in question 8, part c actually. Could somebody explain what was done here exactly? I even checked out the video tutorial in the Examsolutions website based on the question I've asked but I failed to understand this.
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    (Original post by sabahshahed294)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

    Need help in question 8, part c actually. Could somebody explain what was done here exactly? I even checked out the video tutorial in the Examsolutions website based on the question I've asked but I failed to understand this.
    So this is a strange question but doable. OKay so look what you've worked out in part B.And you realise that the left hand side of the equation in part C is the same as part B but without the bottom fraction.

    Knowing this you can just plug in the equation but including the denominator.


    O.s IM sorry im crap at explaining things, put do you kinda get what i mean?
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    (Original post by Dafock)
    So this is a strange question but doable. OKay so look what you've worked out in part B.And you realise that the left hand side of the equation in part C is the same as part B but without the bottom fraction.

    Knowing this you can just plug in the equation but including the denominator.


    O.s IM sorry im crap at explaining things, put do you kinda get what i mean?

    Okay so basically you're saying that the identity given to us from part b needs to be applied into part c, correct? If so, then what to do after that? Cancel like terms? And lol, np on that. I'm bad at it too at times.
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    (Original post by sabahshahed294)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

    Need help in question 8, part c actually. Could somebody explain what was done here exactly? I even checked out the video tutorial in the Examsolutions website based on the question I've asked but I failed to understand this.
    From part c.) you can see that \frac{1+\sqrt{3}tan\theta}{\sqrt  {3}-tan\theta} = tan(\theta +\frac{\pi}{6}), so if you divide both sides by the \sqrt{3}-tan\theta in part c, you can replace the lhs with tan(\theta + \frac{\pi}{6})

    Now when you have tan(theta) = some value you need to take the critical value and the other solutions which are in range, for the tan function, you just add multiples of pi to the critical value and write the answers which are in the range given as 0 to pi.
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    (Original post by sabahshahed294)
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

    Need help in question 8, part c actually. Could somebody explain what was done here exactly? I even checked out the video tutorial in the Examsolutions website based on the question I've asked but I failed to understand this.
    Which part of the written / video solution do you not understand?

    And it will help if you can show us what you tried before getting stuck.
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    (Original post by sabahshahed294)
    Okay so basically you're saying that the identity given to us from part b needs to be applied into part c, correct? If so, then what to do after that? Cancel like terms? And lol, np on that. I'm bad at it too at times.
    OKay think of it this way if i had an equation that was as followed,

    2/x = 1/x + y/x

    you can cancel the x's so.... 2=1 +y

    Thats what they have done they have just cancelled "root3 - tan theta"

    so all you have to do is divide both sides of the equation they gave you by "root3 - tan theta" and you'll see that the left side loos exacty the same as what you did in part b
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    (Original post by NotNotBatman)
    From part c.) you can see that \frac{1+\sqrt{3}tan\theta}{\sqrt  {3}-tan\theta} = tan(\theta +\frac{\pi}{6}), so if you divide both sides by the \sqrt{3}-tan\theta in part c, you can replace the lhs with tan(\theta + \frac{\pi}{6})

    Now when you have tan(theta) = some value you need to take the critical value and the other solutions which are in range, for the tan function, you just add multiples of pi to the critical value and write the answers which are in the range given as 0 to pi.
    So for that, we make a CAST diagram right? (Critical Values) or use the Tan theta graph?
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    (Original post by sabahshahed294)
    So for that, we make a CAST diagram right? (Critical Values) or use the Tan theta graph?
    You could use either method.
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    (Original post by notnek)
    Which part of the written / video solution do you not understand?

    And it will help if you can show us what you tried before getting stuck.
    I didn't really do much with it as I couldn't understand how to proceed with it tbh. Watched the video for that reason but didn't get that as well hence why I asked for help here.
    And Idk why but from my phone, hard to upload photos on TSR
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    (Original post by Dafock)
    OKay think of it this way if i had an equation that was as followed,

    2/x = 1/x + y/x

    you can cancel the x's so.... 2=1 +y

    Thats what they have done they have just cancelled "root3 - tan theta"

    so all you have to do is divide both sides of the equation they gave you by "root3 - tan theta" and you'll see that the left side loos exacty the same as what you did in part b
    Thanks. Understood how to proceed now.

    (Original post by NotNotBatman)
    You could use either method.
    Got it. Thank you.
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    (Original post by sabahshahed294)
    Thanks. Understood how to proceed now.



    Got it. Thank you.
    An alternative method is to use the tan addition formula and solve a quadratic expression in tan theta, from there it might be easier to see that two values of tan theta are needed, as it would be similar to a C2 quadratic expression.
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    (Original post by NotNotBatman)
    An alternative method is to use the tan addition formula and solve a quadratic expression in tan theta, from there it might be easier to see that two values of tan theta are needed, as it would be similar to a C2 quadratic expression.
    Oh alright. I'll try that. Thanks!
 
 
 
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