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Trigonometry identities and equations

  1. Feb 1, 2015 #1
    1) Question statement:
    Simplify 2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x

    2)Relevant equations:
    tan A=sinA/cos A
    1+tan^2A=sec^A
    cot A=1/tanA
    cot A=cos A/sinA
    sin^2A+cos^2A=1
    secA=1/cos A
    cosecA=1/sinA
    1+cosec^2A= cot^2A
    sin2A=2sinAcosA
    cos2A=1-2sin^2A=cos^2A-sin^2A=2cos^A-1
    tan2A=(2tanA)/1-tan^2A

    3) Attempt:
    2sec^2x-2sec^2xsin^2x-sin^2x-cos^2x
    = 2sec^2x(1-sin^2x)-sin^2x-cos^2x
    = 2sec^2x(cos^2x)-sin^2x-cos^2x

    I am stuck after this... Will be helpful if some clues are provided. Thank you!
     
  2. jcsd
  3. Feb 1, 2015 #2

    Dick

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    Science Advisor
    Homework Helper

    Use that sec(x)=1/cos(x).
     
  4. Feb 3, 2015 #3
    Also, at the end you have: (... -sin^2x - cos^2x) . What could you do with that?
     
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