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    Lilp22's Avatar
    Lilp22 Posts: 3, Reputation: 0
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    #1

    Dec 5, 2010, 08:11 AM
    sec^6x-tan^6x = 1 3(tan^2x)(sec^2x)
    We're proving identifies in class, and I was kind confused on how to prove this one.
    Lilp22's Avatar
    Lilp22 Posts: 3, Reputation: 0
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    #2

    Dec 5, 2010, 09:17 AM
    Sorry there was suppose to be a plus in that question
    sec^6x-tan^6x = 1 + 3(tan^2x)(sec^2x)
    neerajtiwari's Avatar
    neerajtiwari Posts: 2, Reputation: 1
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    #3

    Feb 18, 2012, 10:53 PM
    Left Hand Side = sec^6 x - tan^6 x
    = (sec²x - tan²x)(sec^4 x + (tan²x)(sec²x) + tan^4 x)
    = 1*(sec^4 x + (tan²x)(sec²x) + tan^4 x)
    = sec^4 x + (tan²x)(sec²x) + (tan²x)²

    = sec^4 x + (tan²x)(sec²x) + (sec²x - 1)²
    = sec^4 x + (tan²x)(sec²x) + sec^4 x - 2sec²x + 1
    = 1 + 2sec^4 x - 2sec²x + (tan²x)(sec²x)

    = 1 + 2sec²x(sec²x - 1) + (tan²x)(sec²x)
    = 1 + 2(sec²x)(tan²x) + (tan²x)(sec²x)
    = 1 + 3(tan²x)(sec²x) = Right Hand Side
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