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Homework Help: Tough Trig Question

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data
    if tan(x) + cot(x) = 4 then what is

    cos^2(x) + sin^2(x) + tan^2(x) + cot^2(x) + csc^2(x) + sec^2(x)


    2. Relevant equations

    cos^2(x) + sin^2(x) = 1
    cot^2(x) = 1/(tan^2(x))

    3. The attempt at a solution

    = 1 + 14 + 1/(cos^2(x)sin^2(x))

    Now I'm stuck.
     
  2. jcsd
  3. Feb 23, 2010 #2

    Mark44

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    How did you get this? "= 1 + 14 + 1/(cos^2(x)sin^2(x))"
    What's the other side of this equation?
     
  4. Feb 23, 2010 #3
    square first equation and you get tan^2(x) + cot^2(x) +2 = 16
     
  5. Feb 23, 2010 #4

    Mark44

    Staff: Mentor

    You don't "square an equation;" you can square each side of an equation. Are you trying to say that (tan (x) + cot(x))2 = tan2(x) + cot2(x)?

    How about (3 + 4)2? Is that equal to 32 + 42?
     
  6. Feb 23, 2010 #5
    Can we not solve for the first equation for [itex]x[/itex]?

    Express first equation in terms of [itex]\sin(x)[/itex] and [itex]\cos(x)[/itex], get common denominator, and solve for [itex]x[/itex].

    Second equation can be simplified some-what by noting some additional trig identies for [itex]\csc^2(x)[/itex] and [itex]\sec^2(x)[/itex], but we can use solution for [itex]x[/itex] from first equation and substitute to solve for second equation directly.
     
  7. Feb 24, 2010 #6

    Mark44

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    There is no second equation. The idea is to simply evaluate that expression using the value(s) of x obtained from the first (and only) equation.

    The first equation can be solved with minimal substitution, keeping things in terms of tan(x).
     
  8. Feb 24, 2010 #7

    vela

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    No, he's saying

    [tex](\tan x+\cot x)^2=\tan^2 x+2 \tan x\cot x +\cot^2 x=\tan^2 x+2+\cot^2 x[/tex]

    since tan x cot x=1. Because the LHS is also equal to 42, he finds

    [tex]\tan^2 x+\cot^2 x=16-2=14[/tex].
     
  9. Feb 24, 2010 #8

    vela

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    Write this in terms of sin x and cos x and put the LHS over a common denominator. It'll simplify a bit, giving you a result that'll let you finish the problem.
     
  10. Feb 24, 2010 #9

    Mentallic

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    Take a look back at the original equality [itex]tanx+cotx=4[/itex] and note that [itex]tanx=sinx/cosx[/itex] and [itex]cotx=cosx/sinx[/itex] :wink:
     
  11. Feb 24, 2010 #10
    just solved it and I was so close. 1/(cos(x)sin(x)) = 4 Substitute that in for 1/(cos^2(x)sin^2(x)). You get 16. So the answer is 1 + 14 + 16 = 31
     
  12. Feb 24, 2010 #11

    Mentallic

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    Yeah, you were close :tongue:
     
  13. Feb 24, 2010 #12
    Lets not be hasty to bash other's ideas. The cot and tan cancel out, leaving 2.
     
  14. Feb 24, 2010 #13

    Mark44

    Staff: Mentor

    I misread that final +2 on the left side in what Hockeystar posted, probably misreading it as the exponent on cot(x). My mistake.
    "tan^2(x) + cot^2(x) +2 = 16"
     
  15. Feb 24, 2010 #14
    Given: tan(x) + cot (x) = 4

    Square both side then,

    [tex]tan^{2}x + cot^{2}x[/tex] + 2.tan(x) cot(x) = 16

    but tan(x). cot(x) = 1

    therefore, [tex]tan^2{x} + cot^2{x}[/tex] = 14.

    Take this equation first.
     
  16. Feb 24, 2010 #15
    Now again consider tan(x) + cot(x) = 4

    Now tan(x) = [tex]\frac{sin(x)}{cos(x)}[/tex] and similarly cot(x) = [tex]\frac{cos(x)}{sin(x)}[/tex]

    so

    [tex]\frac{sin(x)}{cos(x)}[/tex] + [tex]\frac{cos(x)}{sin(x)}[/tex] = 4

    [tex]\frac{sin^{2}x + cos^{2}x}{sin(x).cos(x)}[/tex] = 4

    then it is equal to

    [tex]\frac{1}{sin(x).cos(x)} = 4[/tex]

    take it equation second.
     
  17. Feb 24, 2010 #16
    Now take your actual equation

    [tex]cos^{2}(x) + sin^{2}(x) + tan^{2}(x) + cot^{2}(x) + csc^{2}(x) + sec^{2}(x)[/tex]

    Now here, [tex]cos^{2}(x) + sin^{2}(x)[/tex] = 1

    [tex]tan^{2}(x) + cot^{2}(x)[/tex] = 14
     
  18. Feb 24, 2010 #17
    now last thing left is [tex]csc^{2}(x) + sec^{2}(x)[/tex]

    [tex]\frac{1}{sin^{2}x}[/tex] + [tex]\frac{1}{cos^{2}x}[/tex]

    that is equal to

    [tex]\frac{cos^{2}x + sin^{2}x}{sin^2(x). cos^{2}x}[/tex]

    now look at equal second and I should leave rest up to you so that you can feel that you have done your home not me.
     
  19. Feb 24, 2010 #18
    now last thing left is [tex]csc^{2}(x) + sec^{2}(x)[/tex]

    [tex]\frac{1}{sin^{2}x}[/tex] + [tex]\frac{1}{cos^{2}x}[/tex]

    that is equal to

    [tex]\frac{cos^{2}x + sin^{2}x}{sin^2(x). cos^{2}x}[/tex]

    now look at equal second and I should leave rest up to you so that you can feel that you have done your home not me.
     
  20. Feb 25, 2010 #19

    Mentallic

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    Oh wow, you gave such a subtle hint there... the OP would definitely feel like the homework was completed entirely on his/her own... :tongue:

    By the way, the OP had all except the [itex]tan^2x+cot^2x[/itex] part figured out before even creating this thread and then did figure out the answer entirely shortly after. You would have known this if you read through the thread.
     
  21. Feb 25, 2010 #20
    Oh! I have seen only the last post and thought the question is still unsolved and so I've given the solution, but I think Mark got an answer that about the last step. Secondly I've not completed this question entirely cause I've seen somewhere in this forum that we have to not give complete solution for any problem. Is I'm right?
     
  22. Feb 25, 2010 #21

    Mentallic

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    Well you basically left a complete solution by stopping near the end where the rest requires no more thinking on the OP's behalf.

    It's like saying and then y=1+14+16 but I won't finish simplifying that because I can't provide a complete solution. You already have.

    But no harm done, the OP already had the answer.
     
  23. Feb 25, 2010 #22
    no, there is an answer in the second equation. He has to only square and add. look
     
  24. Feb 25, 2010 #23
    [tex]
    \frac{1}{sin(x).cos(x)} = 4
    [/tex]

    This is what I've given in the second equation.
     
  25. Feb 25, 2010 #24
    [tex]
    \frac{1}{sin(x).cos(x)} = 4
    [/tex]

    This is what I've given in the second equation.
     
  26. Feb 25, 2010 #25
    how much simpler I should go to prevent myself running agains PF rule. I think I should tell him them do [tex]sin^{2}x + cos^{2}x[/tex] = 1 and then you will be left with [tex]

    \frac{1}{sin^{2}(x).cos^{2}(x)}[/tex] which is equal to [tex]4^{2}[/tex]
     
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