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    Hi guys,
    My maths tutor set me lots of questions to do as part of revision for this week. I'm stuck on one of the questions but she is on holiday so I can't ask her for help.
    So please could someone kindly show me how to do the question in the attachment as I really don't understand it.

    Thanks in advance to anyone to helps
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    (Original post by Aty100)
    Hi guys,
    My maths tutor set me lots of questions to do as part of revision for this week. I'm stuck on one of the questions but she is on holiday so I can't ask her for help.
    So please could someone kindly show me how to do the question in the attachment as I really don't understand it.

    Thanks in advance to anyone to helps
    That tan looks out of place so you want to get rid of that first:

    \displaystyle \frac{t\times s}{1-c} = \frac{\frac{s}{c} \times s}{1-c} = \frac{s^2}{c(1-c)}

    What could you do next?
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    (Original post by notnek)
    That tan looks out of place so you want to get rid of that first:

    \displaystyle \frac{t\times s}{1-c} = \frac{\frac{s}{c} \times s}{1-c} = \frac{s^2}{c(1-c)}

    What could you do next?
    Thank you. That really helped.
    Also I don't completely understand where the 1+ comes from :/
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    (Original post by Aty100)
    Thank you. That really helped. I'm not sure whether I did the last bit correctly. If you don't mind could you please check.
    Also I don't completely under where the 1+ comes from :/
    sin^2(theta) is the same as 1-cos^2(theta). Do you remember the difference of two squares rule?
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    (Original post by OrionMusicNet)
    sin^2(theta) is the same as 1-cos^2(theta). Do you remember the difference of two squares rule?
    I think I got that bit. Not sure if I did it correctly.
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    (Original post by Aty100)
    I think I got that bit. Not sure if I did it correctly.
    It should be

    \displaystyle \frac{1-\cos^2 \theta}{\cos\theta \left(1-\cos \theta \right)}

    You have a negative sign on the denominator. Typo?

    Next step using difference of two squares:

    \displaystyle \frac{(1+\cos \theta)(1-\cos \theta)}{\cos\theta \left(1-\cos \theta \right)}

    Can you see what to do next?
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    (Original post by Aty100)
    I think I got that bit. Not sure if I did it correctly.

    1-c is the same 1 - cos(theta). If you multiply top and bottom of the fraction by cos(theta) you get sin(theta) x sin(theta) on the top, and on the bottom you get cos(theta) - cos^2(theta). You can then factorise the bottom to get cos(theta)(1 - cos(theta)). You can see that if you expand that you would get back to cos(theta) - cos^2(theta). If you change the sin^2(theta) to 1 - cos^2(theta) you have a line in the form of the difference of two squares (a^2 - b^2 is the same as (a-b)(a+b)), therefore 1 - cos^2(theta) is the same as (1 - cos(theta))(1+cos(theta)) as 1^2 is simply 1. Can you see that the (1-cos(theta)) on the top will cancel with the (1-cos(theta) on the bottom of the fraction? You then simply need to split the fraction apart and you should see that you will get the answer.
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    (Original post by notnek)
    It should be

    \displaystyle \frac{1-\cos^2 \theta}{\cos\theta \left(1-\cos \theta \right)}

    You have a negative sign on the denominator. Typo?

    Next step using difference of two squares:

    \displaystyle \frac{(1+\cos \theta)(1-\cos \theta)}{\cos\theta \left(1-\cos \theta \right)}

    Can you see what to do next?
    ohhhh I see. I get it now. Thank you both so much.
 
 
 
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