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Tan^2x =sec^2x-1 also: tan= sec -1 or am I missing something?

  1. Feb 17, 2015 #1
    A particular problem with factoring has both of these, one in the denominator and one in the numerator, if it were algebra it would look like: x^2-1/x-1. The trouble is I've forgotten how to simplify this. I'm on taptalk.)
     
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  3. Feb 17, 2015 #2

    Mentallic

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    Can you please make more sense of this. What is your problem exactly? And no, [itex]\tan{x}\neq \sec{x}-1[/itex]
     
  4. Feb 17, 2015 #3
    That just looks like gibberish on taptalk. Essentially it is Sec^2x-1/Secx-1 once a bit of factoring is done.
     
  5. Feb 17, 2015 #4

    Mentallic

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    So you want to simplify this expression by factoring or any other means? Ok well, if we take a look at

    [tex]x^2-\frac{1}{x}-1[/tex]

    and then factor out 1/x giving us

    [tex]\frac{1}{x}\left(x^3-x-1\right)[/tex]

    The cubic has no rational factors, so that is the best we can do. We're not completely at a loss with simplifying though.

    You know that

    [tex]\tan^2{x}=\sec^2{x}-1[/tex]

    so then use this to simplify your expression.
     
  6. Feb 17, 2015 #5
    ImageUploadedByTapatalk1424221100.710027.jpg
    This is my problem.
     
  7. Feb 17, 2015 #6
    I solved my problem... I was not removing the square when factoring. Always seems to be that kind of mistake that gets me.
     
  8. Feb 17, 2015 #7

    Mentallic

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    Please use parentheses in future.

    x^2-1/x-1
    is read as
    [tex]x^2-\frac{1}{x}-1[/tex]

    while
    (x^2-1)/(x-1)
    [tex]=\frac{x^2-1}{x-1}[/tex]
     
  9. Feb 19, 2015 #8

    HallsofIvy

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    [tex]\frac{x^2- 1}{x- 1}= \frac{(x- 1)(x+ 1)}{x- 1}= x+ 1[/tex] as long as x is not equal to 1.
     
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