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Imta 2009 Q6

  • 01-03-2012 05:53PM
    #1
    shaunie007
    Registered Users, Registered Users 2 Posts: 79 ✭✭


    Howdy,

    I'm looking for the solution to this problem please:

    http://www.imta.ie/competitions.htm

    Click on the 2009 comp. Round 3, Q6.

    Cheers


Comments

  • #2
    Oween
    Registered Users, Registered Users 2 Posts: 16 ✭Oween


    Howdy Shaunie007,

    Let 〖(sec〗⁡〖θ)〗^2=x/(x-1)
    ⇒sec⁡〖θ=√x/√(x-1)〗
    〖⇒cos〗⁡〖θ=√(x-1)/√x 〗 (From triangle)
    〖⇒sin〗⁡〖θ=1/√x (From triangle)
    〖〖⇒(sin〗⁡θ)〗^2=1/x
    ∴f((〖sec⁡θ)〗^2 )=(sin⁡〖θ)〗^2

    Any questions feel free to ask...


  • #3
    Oween
    Registered Users, Registered Users 2 Posts: 16 ✭Oween


    Let (secθ)^2=x/(x-1)
    ⇒secθ=√x/√(x-1)
    ⇒cosθ=√(x-1)/√x (From Triangle)
    ⇒sinθ=1/√x (From Triangle)
    ⇒(sinθ)^2=1/x
    ∴f((secθ)^2 )=(sinθ)^2
    Sorry, this might be a bit clearer...


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