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March 28, 2017

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secx/secx-tanx=sec^2x+ secx+tanx

  • precal- PLEASE HELP! - ,

    suspect you might mean

    sec x / (sec x -tan x) = sec^2 x + sec x + tan x
    work on the left
    1/cos x / [ 1/cos x -sin x/cos x ] =

    1/[1 - sin x] =
    [ 1 + sin x ]/ [1-sin^2 x] =
    [1+sin x ]/cos^2 x =

    work on the right
    = 1/cos^2 x + 1/cos x + sin x/cos x)
    = (1/cos^2 x) [1 + cos x + sin x cos x]
    sorry, I can not make that come out. Please check for typos

  • precal- PLEASE HELP! - ,

    Sorry, it was a typo, it's supposed to be:
    sec/(secx-tanx)=sec^2x+secxtanx

  • precal- PLEASE HELP! - ,

    LS = (1/cosx)/(1/cosx - sinx/cosx)
    = (1/cosx)/( (1-sinx)/cox )
    = (1/cosx)(cosx)/(1-sinx)
    = 1/(1-sinx)

    RS = 1/cos^2 x + (1/cosx)(sinx/cosx)
    = 1/cos^2 x + sinx/cos^2 x
    = (1+ sinx)/cos^2x

    back to LS
    LS = 1/(1-sinx) * (1+sinx)/(1+sinx)
    = (1+sinx)(1 - sin^2 x)
    = (1+sinx)/cos^2 x
    = RS

  • precal- PLEASE HELP! - ,

    Well then we are finished
    work on the right
    = 1/cos^2 x + (1/cos x * sin x/cos x)
    = (1/cos^2 x) [1 + sin x]

  • precal- PLEASE HELP! - ,

    :)thanks!!! I've been trying to figure this out for a while now, can you explain this to me: what do I do when i get to something like this
    (cosx/sinx-1/(sinx))?

  • precal- PLEASE HELP! - ,

    If you see (sin x - 1) on the bottom try multiplying top and bottom by (sin x + 1)
    like
    1/(sin x - 1) * [ (sin x + 1)/(sin x+1) ]

    = (sin x + 1) / (sin^2 x - 1)

    = (sin x +1)/-cos^2 x

  • precal- PLEASE HELP! - ,

    thanks alot!!!!!!!!!!!!!!

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