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Trig questions

  1. Feb 27, 2006 #1
    i need help for these 2 trig proofs, i did everything i could but it's impossible.

    1st question; (cot^2X)-1=csc^2X


    2nd question; (cot^2X)-(cos^2X)=cos^2Xcot^2X

    caution, both might be insoluable

  2. jcsd
  3. Feb 27, 2006 #2
    You can use the Pythagorean Identities to evaluate the expressions to see if they are equal. Recall that [tex]\cos^{2} \theta + \sin^{2} \theta = 1[/tex] and that you can modify this equation by dividing by sin or cos to give two additional equations in terms of tan, cot, sec, and csc.
  4. Feb 27, 2006 #3


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    Science Advisor

    Do you know the identity sin^2(x) + cos^2(x) = 1? Try dividing, rearranging terms, etc.
  5. Feb 27, 2006 #4
    i tried everything but it does not work at all!!!!!!!!!!!!!!!!
  6. Feb 27, 2006 #5
    You need to use the identity we've given you. Divide the identity by sin x and see what you come up with. You should have something in terms of cotangent and cosecent.
  7. Feb 27, 2006 #6
    oh i forgot tell you that my ass hole teacher want to us to do the proof by solving either side
    so i can not divide, square, multiply or anything to the both side or the one side
  8. Feb 27, 2006 #7
    I know and what I'm trying to get you to do is complete this one step so you can compare the result to your first question. What do you get when you divide the identity above by sin x?
  9. Feb 28, 2006 #8


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    Homework Helper

    I am going to give you an example, which is nearly the same as this one:
    Prove that:
    sec2x = tan2x + 1
    I am going from the LHS to the RHS:
    [tex]\sec ^ 2 x = \frac{1}{\cos ^ 2 x} = \frac{\sin ^ 2 x + \cos ^ 2 x}{\cos ^ 2 x} = \tan ^ 2 x + 1[/tex] (Q.E.D)
    By the way, your first problem is not correct, it should read:
    cot2x + 1 = csc2x
    cot2x - 1 = csc2x
    For the second problem, what's cot2x in terms of sin(x), and cos(x)? You should also note that:
    sin2x + cos2x = 1 (the Pythagorean Identity)
    Can you go from here? :)
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