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Solving Trig Identities.

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    1/cscx-sinx = secx tanx

    2. Relevant equations

    cscx = 1/sinx
    secx = 1/cosx

    3. The attempt at a solution

    1/cscx-sinx = secx tanx

    L.S.
    = 1/cscx-sinx
    = 1/(1/sinx)-sinx

    R.S.
    = secx tanx
    = (1/cosx)(sinx/cosx)

    This is where I'm getting confused. Why can't I make the L.S equal the Right side?
     
  2. jcsd
  3. Apr 26, 2010 #2

    rock.freak667

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    Start with one side alone and make that match the other side.


    What happens if you multiply both the numerator and denominator by sinx/sinx ?
     
  4. Apr 26, 2010 #3
    (1/sinx)(sinx/sinx) = sin^2x - sinx?
     
  5. Apr 26, 2010 #4

    rock.freak667

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    [tex]\frac{1}{\frac{1}{sinx}-sinx} \times \frac{sinx}{sinx}[/tex]

    Redo it.
     
  6. Apr 26, 2010 #5
    Hmmm. Do I multiply everything in the first denominator by sinx? Or just the - sinx?

    I get sinx/sinx-sin2x. I think... I've never seen mulitiplication like this. It's probably something easy, I've just never done it yet.
     
  7. Apr 26, 2010 #6

    rock.freak667

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    Multiply everything in the numerator by sinx, and multiply everything in the denominator by sinx.
     
  8. Apr 26, 2010 #7
    ok so I get sinx/(sinx/sin^2x)-sin^2x

    If I divide that I end up with sinx-sinx = 0?
     
  9. Apr 26, 2010 #8
    err or would it be sinx/sinx-sinx?
     
  10. Apr 26, 2010 #9
    If that's the actual problem(can't tell from your original post), then:

    get a common denominator:

    1/((1-((sin x)^2))/ sin x) becomes:

    (sin x)/(1-((sin x)^2)) --> 1 - sin^2 x = cos^2 x:

    (sin x)/((cos x)^2) = sec x tan x
     
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