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Trig Proof

  1. Jul 5, 2005 #1
    yea ok this problem has been making me mad for the past 2 weeks or so (it's a summer project due next week) any help would b great

    sqrt of ((1-sinx)/(1+sinx))= |secx-tanx|

    i tried:
    and i need it to equal the other side without touching the other side... please HELP!!!
  2. jcsd
  3. Jul 5, 2005 #2


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    Hint: [itex]|A| = \sqrt{A^2}[/itex]
    Hint: [itex]\cos ^2x + \sin ^2x = 1[/itex]
  4. Jul 5, 2005 #3
    sorry but that still seems to not work, wat can i do? am i doing something wrong?
  5. Jul 5, 2005 #4


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    on the sqrt side you could multiply the numerator and denominator by 1-sin(x)
    and use (sin(x))^2=1-(cos(x))^2
    on the other side write it in terms of sin and cos as you have but then writ cos in terms of sin (hit use sqrt).
  6. Jul 5, 2005 #5
    it works! thanks
  7. Jul 5, 2005 #6


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    You must have been doing something wrong, because it does indeed work:

    [tex]\left |\frac{1-\sin x}{\cos x}\right | = \sqrt{\frac{(1-\sin x)^2}{\cos ^2 x}} = \sqrt{\frac{(1-\sin x)^2}{1 - \sin ^2 x}} = \sqrt{\frac{(1-\sin x)^2}{(1 - \sin x)(1 + \sin x)}} = \sqrt{\frac{1-\sin x}{1 + \sin x}}[/tex]

    Note that I cancelled a factor of [itex](1 - \sin x)[/itex] without checking that it wasn't 0. But you can check for yourself that if it is zero, then sin(x) = 1, which implies that cos(x) = 0, and so sec(x) and tan(x) are undefined, and I would assume that you're only asked to show that this identity holds for those values of x that don't lead to us having anything undefined (this happens when sec(x) or tan(x) are undefined, or when (1 + sin(x)) = 0).
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