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    show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50

    cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)

    = -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2x

    don't know where to go from there
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    (Original post by AndyOC)
    show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)= -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2xdon't know where to go from there
    I don't understand how you've written the question? Is it cosecx divided by (1+cosecx) - cosecx divided by (1-cosecx)? :s
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    (Original post by Save.Me)
    I don't understand how you've written the question? Is it cosecx divided by (1+cosecx) - cosecx divided by (1-cosecx)? :s
    yeah that's it, sorry it's not clear
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    (Original post by AndyOC)
    show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50

    cosecx (1 - cosecx) - cosecx( 1 + cosecx) / (1 + cosecx)( 1 - cosecx)

    = -2cosec^2x / 1 - cosec^2x = -2cosec^2x / -cot^2x

    don't know where to go from there
    Make them all into sine and cos e.g. 1-(1/sinx) = (sinx-1)/sinx
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    just replace cosecx with a letter ... w/( 1 + w ) - w/ ( 1 - w ) = 50 and just do normal algebra ?
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    You have caused a lot of confusion. the actual question is AQA MPC3 Jan 2011, Q. 7b

    show that equation cosecx / (1 + cosecx) - cosecx / (1 - cosecx) = 50


    can be written in the form sec^2x = 25

    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
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    (Original post by kkboyk)
    Make them all into sine and cos e.g. 1-(1/sinx) = (sinx-1)/sinx
    (Original post by the bear)
    just replace cosecx with a letter ... w/( 1 + w ) - w/ ( 1 - w ) = 50 and just do normal algebra ?
    Yeah I've done it now anyway, just turned cosec^2 into sin fraction and cot^2 into tan fraction then divided. Thanks for the help
 
 
 
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