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How do you prove this Trig identity?

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    (tan/(1-cot))+(cot/(1-tan))=1+(sec)(csc)

    3. The attempt at a solution

    This problem showed up in my class as a warm-up, and when my teacher tried to solve it for us he got stuck. I've asked a few classmates in higher level math, and they all seem to get stuck.

    I myself have tried to prove the identity. I managed to simplify the right side down to (tan*cot)+(tan*cot)/(sin*cos) but I get stuck after this.

    Any help is much appreciated.

    Thanks!
     
  2. jcsd
  3. Mar 19, 2012 #2
    This problem was fun to solve! :)

    In my opinion, there are a couple of ways to tackle these equations:

    Start with the more complicated side, reduce EVERYTHING to sines and cosines, combine into one fraction, and simplify.
    Remember that [itex]1 = sin^2(x) + cos^2(x)[/itex] and a few other identities (half angle formulas, etc), some problems involve clever use of these.
    Remember to always start with one side and show it is equal to the other side, never start with both sides equal to each other.


    So, for this problem, I would recommend starting with the left side and converting everything to sines and cosines (it gets messy!). Eventually you will see a difference of cubes in the numerator, that is when you know you are on the right track!
     
  4. Mar 19, 2012 #3

    AlephZero

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    A good way to start with this type of identity is to write everything using just sin and cos.

    tan/(1-cot)
    = (sin/cos) / (1 - cos/sin)
    = (sin/cos) / ((sin - cos) / sin)
    = sin2 / (cos(sin - cos))

    Do the same sort of thing for the other term on the left hand side, and also for the right hand side.
     
  5. Mar 22, 2012 #4
    It's been a few days since first seeing this problem.

    I asked another math teacher, and he literally spent an entire class trying to solve it. My teacher has asked other teachers, and they haven't figured it out yet.

    I've worked the left side down to:
    ((sin^2)/((cos)(sin-cos)))+((cos^2)/((sin)(cos-sin)),
    but I don't know what to do from here. I'm not sure how to simplify the denominators properly.

    Thanks
     
  6. Mar 22, 2012 #5
    Good start! That's exactly where you want to be! Now your next step is to combine the fraction into one! This is where it starts to get tricky, and once your combine your fraction you may not see it at first. Notice my hint in my previous post: you will have a difference of cubes in the numerator. You need to eventually factor that. Notice that in one denominator you have [itex]cos(x)[sin(x) - cos(x)][/itex] and the other you have [itex]sin(x)[cos(x) - sin(x)][/itex]. What would the common denominator be in order that you can combine the two fractions into one? Be careful about your signs! :)

    Edit: I'm putting this fraction here so it's easier to visualize than reading it as one line of text. Here is what you have so far:

    [itex]\displaystyle\frac{sin^2(x)}{cos(x)[sin(x) - cos(x)]} + \frac{cos^2(x)}{sin(x)[cos(x) - sin(x)]}[/itex]
     
    Last edited: Mar 22, 2012
  7. Mar 22, 2012 #6
    I get

    ((sin^3(cos-sin)+cos^3(sin-cos))/((sin)((cos-sin)(cos)(sin-cos))

    but somehow this doesn't seem right.
    How do I get a difference of squares?
     
  8. Mar 22, 2012 #7
    What you did certainly wasn't wrong, but there's a more compact way of writing it. Try starting here:

    [itex]\displaystyle\frac{sin^2(x)}{cos(x)[sin(x) - cos(x)]} + \frac{cos^2(x)}{sin(x)[cos(x) - sin(x)]} = \frac{sin^2(x)}{cos(x)[sin(x) - cos(x)]} - \frac{cos^2(x)}{sin(x)[sin(x) - cos(x)]}[/itex]

    Do you see how the minus sign got there? Can you combine them into one fraction now?
     
  9. Mar 22, 2012 #8
    somewhere along the way you will need the ( A - B)^3 formula.
     
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