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Help on trig identity

  1. Apr 14, 2009 #1
    Help plz on trig identity

    1. The problem statement, all variables and given/known data
    Simplify cot2xsecx + 1/cosx





    3. The attempt at a solution
    Well so far i got:
    cot2xsecx + 1/cosx
    =(cos2x/sin2x)(1/cosx) + 1/cosx
    =((1+cos2x)/(1-cos2x))(1/cosx) + 1/cosx


    and from there im stuck, I've tried playing around with it but I just seem to get to dead ends.
    any help is appreciated, Thanks
     
    Last edited: Apr 14, 2009
  2. jcsd
  3. Apr 14, 2009 #2

    Cyosis

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    Re: Help plz on trig identity

    In your second step cancel the square of the cosine. Then merge the entire term into one fraction.
     
  4. Apr 14, 2009 #3
    Re: Help plz on trig identity

    wow that was fast, thanks

    so by cancelling the square I get:
    (cosx/sinx)(1/cosx) + 1/cosx
    =sinx/cos2x + 1/cosx <--- multiply by cosx
    =sinx/cos2x + cosx/cos2x
    =sinx + cosx

    did i do that correctly?
    the question im doing is multiple choice answer tho
    a. csc2xsecx
    b. sec3x
    c. sec2xcscx
    d. csc3x
     
  5. Apr 14, 2009 #4
    Re: Help plz on trig identity

    [tex] \cot^2{x} \sec{x} + {1 \over \cos{x}} [/tex]

    = [tex] \cot^2{x}\sec{x} + \sec{x} [/tex]

    = [tex](\cot^2{x} + 1) \sec{x} [/tex]

    = [tex]\csc^2 {x} \sec{x} [/tex]

    If you are familiar with the identity, [tex] \cot^2{x} + 1 = \csc^2 {x} [/tex], it's obvious.

    EDIT: In case you aren't familiar with that particular identity, it's simple to show with a couple more steps:
    [tex] 1 + cot^2{u} [/tex]

    =[tex]1 + {cos^2{u} \over sin^2{u}} [/tex]

    =[tex]{sin^2{u} + cos^2{u} \over sin^2{u}} [/tex]

    =[tex]{1 \over \sin^2{u} } [/tex]

    =[tex]\csc^2{u} [/tex]
     
    Last edited: Apr 14, 2009
  6. Apr 14, 2009 #5
    Re: Help plz on trig identity

    Also, in you solution, this step is not valid:

     
  7. Apr 14, 2009 #6
    Re: Help plz on trig identity

    awsome..thanks guys, omfg i forgot bout that identity
     
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