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    im stuck on proving an identity!
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    i have the answer....but i dont know how to type it all out...its too long. i'll try scanning it or something....hold on k? may take a while!
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    (Original post by IntegralAnomaly)
    im stuck on proving an identity!
    I'm actually quite shocked, that you actually need help.
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    (Original post by Bhaal85)
    I'm actually quite shocked, that you actually need help.
    ? I always need help,thats how i learn
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    (Original post by IntegralAnomaly)
    ? I always need help,thats how i learn
    Is this like P5 or P6 level?
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    (Original post by Bhaal85)
    Is this like P5 or P6 level?
    edexcel p6.
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    (Original post by IntegralAnomaly)
    edexcel p6.
    Sorry, I can't help you, I'm on the OCR board and have only gone up to P3.
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    (Original post by Bhaal85)
    Sorry, I can't help you, I'm on the OCR board and have only gone up to P3.
    Its oke,thanks anyway.
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    Have u asked on Nrich?
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    (Original post by JamesF)
    Have u asked on Nrich?
    no.I don't think they reply that much on there nowadays,maybe due to exams.
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    there's supposed to be side working where there is a purple star.

    * tan(A-B)= tanA-tanB/(1+tanAtanB)
    therefore 1 + tanAtanB= tanA-tanB/tan(A-B)

    hope this helps!
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    okay so you have got:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

    Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

    Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

    Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

    Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
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    (Original post by p8224)
    there's supposed to be side working where there is a purple star.

    * tan(A-B)= tanA-tanB/(1+tanAtanB)
    therefore 1 + tanAtanB= tanA-tanB/tan(A-B)

    hope this helps!
    thanks.
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    (Original post by theone)
    okay so you have got:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

    Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

    Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

    Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

    Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
    and thanks to u 2.
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    (Original post by IntegralAnomaly)
    im stuck on proving an identity!
    one word.
    yuck.
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    (Original post by kimoni)
    one word.
    yuck.
    u mean beautiful.
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    (Original post by IntegralAnomaly)
    u mean beautiful.
    only when you know how!!
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    (Original post by theone)
    okay so you have got:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1.

    Now consider tan(k+1)x cot(x) + 1 = (tan(k+1)x + tan(x))/tan(x).

    Now tan(k+1)x+tan(x) = tan(k+2)x (1-tan(x)tan(k+1)x).

    Hence (tan(k+1)x + tan(x))/tan(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x.

    Now subtituting tan(k+1)x cot(x) = tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 gives:

    tan(k+1)x tan(k+2)x + tan(k+1)x cotx - k - 1. = tan(k+1)xtan(k+2)x + tan(k+2)xcotx - tan(k+2)xtan(k+1)x - 1 - k - 1 = tan(k+2)xcot(x) - (k+1) - 1.
    I was about to say that.
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    (Original post by Bhaal85)
    I was about to say that.
    yea,and i knew that but i was just testing u people
 
 
 
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