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    Realised I struggle with this topic quite a bit and probably will continue to do so, so might as well use one thread for it.

    First question I'm stuck on:

    "By letting tanx=t, show that the equation 4tan2x + 3cotx sec^(2)x = 0 becomes (3t)^4-(8t)^2-3=0"

    I split each bit of the first equation into parts to try and get them in terms of t.
    4tan2x became 8t/(1-(t)^2) using the double angle formula
    3cotx = 3/t as I know cot = 1/tan
    Couldn't get sec^2x in terms of t, not sure how to proceed.
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    1+tan^2x = sec^2x

    (Original post by DarkEnergy)
    Realised I struggle with this topic quite a bit and probably will continue to do so, so might as well use one thread for it.

    First question I'm stuck on:

    "By letting tanx=t, show that the equation 4tan2x + 3cotx sec^(2)x = 0 becomes (3t)^4-(8t)^2-3=0"

    I split each bit of the first equation into parts to try and get them in terms of t.
    4tan2x became 8t/(1-(t)^2) using the double angle formula
    3cotx = 3/t as I know cot = 1/tan
    Couldn't get sec^2x in terms of t, not sure how to proceed.
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    (Original post by NotNotBatman)
    1+tan^2x = sec^2x
    Thanks, haven't been taught that.
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    (Original post by DarkEnergy)
    Thanks, haven't been taught that.
    It comes from sin^2x+cos^2x=1

    dividing through by cos^2x gives :  \frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

    It follows,  tan^2x + 1 = sec^2x

    Try dividing by sin^2x and you'll get another identity.
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    (Original post by NotNotBatman)
    It comes from sin^2x+cos^2x=1

    dividing through by cos^2x gives :  \frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

    It follows,  tan^2x + 1 = sec^2x

    Try dividing by sin^2x and you'll get another identity.
    cosec^2x=cot^2x+1 ?
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    (Original post by DarkEnergy)
    cosec^2x=cot^2x+1 ?
    yes.
 
 
 
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