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    I've seen 2 questions in the text book but i cannot do them!
    You have to do something like the left hand side = and then make it come up as the right hand side:confused:
    they are:

    1) sec²θ(cot²θ-cos²θ ) ≡cot²θ

    2)( sin2θ+sinθ )/( 1+cos2θ+cosθ )≡tanθ

    Any ideas??
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    are you revising c3-retake? lol good boy.
    hopefully no retake for me..

    i think you need to take L.H.S and use the trignometric identities try it and think about it.
    don't expand the bracket! just try simplfying what is inside the bracket.
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    no... just started doing it!
    doin whole a level in 1 year
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    (Original post by notan123)
    no... just started doing it!
    doin whole a level in 1 year
    WoW! thats intensive man! i hope you get the grades you're after
    im just starting c4 next week lol
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    Re-write sec²θ for question 1.
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    then expand with the brackets ?
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    and Re-write sin2θ and cos2θ for question two
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    (Original post by notan123)
    then expand with the brackets ?
    Try and see what happens.
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    1.) LHS:
    sec²θ(cot²θ-cos²θ ) ≡cot²θ
    1/cos²θ x (cos²θ /sin²θ - cos²θ )
    cos²θ /sin²θcos²θ - cos²θ/cos²θ
    1/sin²θ – 1
    cosec²θ -1 = cot²θ


    using identity: sin²θ + cos²θ = 1
    Divide by sin²θ would give:
    sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
    1 + cot²θ = cosec ²θ
    cosec²θ – 1 = cot²θ
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    (Original post by nushii)
    1.) LHS:
    sec²θ(cot²θ-cos²θ ) ≡cot²θ
    1/cos²θ x (cos²θ /sin²θ - cos²θ)
    cos²θ /sin²θcos²θ - cos²θ/cos²θ
    1/sin²θ – 1
    cosec²θ -1 = cot²θ


    using identity: sin²θ + cos²θ = 1
    Divide by sin²θ would give:
    sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
    1 + cot²θ = cosec ²θ
    cosec²θ – 1 = cot²θ
    Read the rules Mr.

    anyway, he said "help" not "answer"
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    (Original post by nushii)
    1.) LHS:
    sec²θ(cot²θ-cos²θ ) ≡cot²θ
    1/cos²θ x (cos²θ /sin²θ - cos²θ)
    cos²θ /sin²θcos²θ - cos²θ/cos²θ
    1/sin²θ – 1
    cosec²θ -1 = cot²θ


    using identity: sin²θ + cos²θ = 1
    Divide by sin²θ would give:
    sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
    1 + cot²θ = cosec ²θ
    cosec²θ – 1 = cot²θ
    Read the rules Ms.

    anyway, he said "help" not "answer"
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    (Original post by 12ian34)
    Read the rules Mr.

    anyway, he said "help" not "answer"
    Im a girl and i didnt read it properly Sorry
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    for q2 would i write sin2θ as 2sinθcosθ and cos2θ as cos²θ - sin²θ
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    (Original post by notan123)
    for q2 would i write sin2θ as 2sinθcosθ and cos2θ as cos²θ - sin²θ
    You're on the right track with the sin2θ, but cos2θ should be written in one of the other ways. There are three ways to expand cos2θ.

    This site might help a bit more, but only look if you can't think of another way:

    http://library.thinkquest.org/20991/alg2/trigi.html

    (Original post by nushii)
    Im a girl and i didnt read it properly Sorry
    Ha sorry about that, I edited it about 30 seconds after I posted that comment once i noticed!
    Also, about the rules, I mean, you shouldn't really give the whole answer, just nudge them so they figure it out themselves, but don't worry about it!
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    (Original post by notan123)
    no... just started doing it!
    doin whole a level in 1 year
    Me too! Well, I'm doing 3 but maths' the nicest (minus D1 that is!)
 
 
 
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