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Pre calc proof

  1. Jan 31, 2006 #1
    ok i have been working on this problem for a little bit and am stumped. i need to prove.
    (sec^2-6*tan+7)/sec^2-5 = (tan-4)/tan+2

    this is what i have done skipping a few steps
    ((1/u^2)-6*(v/u)+7)/(1-5u^2)/u^2

    then (u-6u^3*vu^2*1-5u^3)/u^3 got this by finding a common denominator and flipping the bottom fraction. i missed quite a bit of class just trying to catch up but have no idea what i am doing
     
  2. jcsd
  3. Jan 31, 2006 #2

    Well firstly why are you making substitutions like that, there is absolutely no reason too and makes it incredibly unobvious when you use a trig identity if you do that. Second your notation is mostly gibberish to me, I have no clue what "tan-4) or "tan+2" are supposed to be.
     
  4. Jan 31, 2006 #3
    that was the way i was taught.
    sec= secant=1/u
    tan= tangent=u/v

    so tan-4 would be (u/v)- 2
     
  5. Jan 31, 2006 #4

    VietDao29

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    As d_leet already said, tan + 4, tan - 2 really make no sense.
    You should write tan(x) + 4, or tan(x) - 2 instead.
    back to the problem, I hope you mean:
    Prove:
    [tex]\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan (x) - 4}{\tan (x) + 2}[/tex]
    Is that what you mean?
    I'll give you a hint, try to change all secant function to tangent function. Do you know:
    [tex]1 + \tan ^ 2 (x) = \sec ^ 2 (x)[/tex]?
    Then factor it, and shortly arrive at what you want to prove. :smile:
     
  6. Jan 31, 2006 #5
    yes that is what i mean i will try that
     
  7. Jan 31, 2006 #6
    That substitution is one of two things, either completely wrong, or very unusual, put everything in terms of sine and cosine, that might help better, and I still have no clue what "tan-4" means.

    Because if secant is 1/u, thangent cannot be u/v it would be v/u assuming that v = sine and u = cosine.
     
  8. Jan 31, 2006 #7

    VietDao29

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    Nooooo, you again forget the angle, the tangent, secant values of what angle? (x, y, t, k, [itex]\alpha, \ \beta, \ \gamma, \ \theta[/itex] or what?)
    ------------
    EDIT: Have you solved the problem? :smile:
     
    Last edited: Jan 31, 2006
  9. Jan 31, 2006 #8
    i relised that after i posted it is v/u also that is what i used in the first problem. and no i have not solved it yet i think not taking algebra 1 or 2 is hurting me now
     
  10. Jan 31, 2006 #9
    ok i dont know if you can do this if so why did my answer come out flipped
    1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 to
    tan^2(x) -5 tan(x) +7/ tan^2(x) -4 to
    tan(x) -5 tan(x) +7/ tan(x) -4 to
    tan(x) +2/ tan (x) -4
     
  11. Jan 31, 2006 #10
    1+ tan^2(x) -6 tan(x) +7/ 1+ tan^2(x) -5 this is what i get after changing them to tangents
    i remember doing quadratic equation once but dont remember what they were and i thought i could factor but cant see it in this problem. got any links handy that can help with this problem

    edit there was a post above mine
    i think that by factoring you want me to see that 1 + tan^2(x) can be factored like a^2+b^2 but i cant remember if there are others that apply to this problem
     
    Last edited: Jan 31, 2006
  12. Jan 31, 2006 #11
    Combine like terms and you should find that those expressions factor very nicely.
     
  13. Jan 31, 2006 #12

    VietDao29

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    Arrghh, you should take Algebra before taking pre-calculus!!! i.e, you should have some solid, and basic knowledge of algebra, or you'll not learn much in pre-calculus. You can either find some Algebra book and study it by yourself, or you can ask a tutor... (a tutor maybe better).
    ---------------------
    You know what addition and multiplication are, right?
    Commutative property of addition: a + b = b + a, ie two numbers add to the same thing whichever order you add them in.
    Example:
    2 + 3 = 5
    3 + 2 = 5
    Commutative property of multiplication: ab = ba
    3 . 5 = 15
    5 . 3 = 15
    Associative property of addition: (a + b) + c = a + (b + c), ie, if you take a + b first, then add c, it's the same as you add b + c together first, and then add a.
    Example:
    (1 + 9) + 7 = 10 + 7 = 17
    1 + (9 + 7) = 1 + 16 = 17.
    Associative property of multiplication: (ab)c = a(bc).
    Example:
    (2 . 3) . 5 = 6 . 5 = 30
    2 . (3 . 5) = 2 . 15 = 30
    Distributive property of multiplication with respect to addition: a(b + c) = ab + ac
    Example:
    2 . (3 + 4) = 2 . 7 = 14
    2 . 3 + 2 . 4 = 6 + 8 = 14
    Additive inverse, or opposite, of a number n is the number which, when added to n, yields zero. We denote it to be: -n.
    That means n + (-n) = 0
    Example:
    The additive inverse of 7 is −7, because 7 + (−7) = 0
    The additive inverse of a is −a (definition).
    The additive inverse of (ab) = -(ab).
    Subtraction is the reverse of addition, to subtract b from a, we do as follow:
    a - b = a + (-b), ie, we add a and the opposite number of b.
    -----------------------
    Say, you want to simplify the expression:
    3x - 7y + 12x - 4y - 9y = 3x + (-7y) + 12x + (-4y) + (-9y)
    = 3x + 12x + (-4y) + (-9y) + (-7y) (Associative property of addition)
    = (3 + 12)x + ((-4) + (-9) + (-7))y = 15x + (-20)y = 15x - 20y.
    Do the same, can you go from
    [tex]\frac{\tan ^ 2 (x) + 1 - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5}[/tex] to [tex]\frac{\tan ^ 2 (x) - 6 \tan (x) + 8}{\tan ^ 2 (x) - 4}[/tex]?
    -----------------------
    QUADRATIC EQUATION:
    Look here for quadratic equations.
    -----------------------
    FACTORIZATION:
    See here for some basic stuff about factorization.
    -----------------------
    Remember that, once we've factor both numerator and denominator, we can cancel out something that both numerator, and denominator have.
    Example:
    [tex]\frac{(x - 2) (3x + 5) (x + 7)}{(6x - 5) (x - 2) (x + 7)}[/tex]
    (x - 2) and (x + 7) are in both numerator, and denominator, cancelling them out, we have:
    [tex]\frac{(x - 2) (3x + 5) (x + 7)}{(6x - 5) (x - 2) (x + 7)} = \frac{3x + 5}{6x - 5}[/tex]
    Can you do the problem now?
    My last advice is: Go buying some book, and study Algebra, or hire some good tutor.
    GET ALGEBRA BOOK! :smile:
     
    Last edited: Jan 31, 2006
  14. Feb 10, 2006 #13
    ok i understand this what i dont is how tan(x)+2/tan(x)-4 = tan(x)-4/tan(x)+2
     
  15. Feb 11, 2006 #14

    VietDao29

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    No, they are not equal:
    [tex]\frac{\tan x + 2}{\tan x - 4} \neq \frac{\tan x - 4}{\tan x + 2}[/tex]
    Since, it's quite a long time, and you may not have worked out the problem, I think I can post the solution.
    Prove that:
    [tex]\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan (x) - 4}{\tan (x) + 2}[/tex]
    ---------
    Proof:
    We will now prove that the LHS of the equation is equal to the RHS. So we will start from the LHS:
    [tex]\frac{\sec ^ 2 (x) - 6 \tan (x) + 7}{\sec ^ 2 (x) - 5} = \frac{\tan ^ 2 x + 1 - 6 \tan x + 7}{\tan ^ 2 x + 1 - 5} = \frac{\tan ^ 2 x - 6 \tan x + 8}{\tan ^ 2 x - 4}[/tex]
    We then factor both numerator and denominator:
    [tex]\frac{(\tan x - 2) (\tan x - 4)}{(\tan x - 2) (\tan x + 2)} = \frac{\tan x - 4}{\tan x + 2}[/tex] (Q.E.D)
    ---------
    Can you understand this proof? And have you got an Algebra book yet? :)
     
  16. Feb 15, 2006 #15
    Oh man :cry:

    We all have trouble sometimes :)

    I have been working on a long homework assignment for the past 6 hours, I wish I had never taken thermodynamics
    :cry: :cry:
     
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