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Simplify Trigonometry expression

  1. Jul 10, 2016 #1
    1. The problem statement, all variables and given/known data
    I want to know how he simplified this expression to tan (theta)
    Untitled.jpg

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 10, 2016 #2
    Hint: How is ##\tan \theta## related to ##\sin \theta## and ##\cos \theta##?
     
  4. Jul 10, 2016 #3
    I know it's Sin/Cos however, I don't know how he did it.
     
  5. Jul 10, 2016 #4
    So how do you get ##\sin \theta / \cos \theta## from the original expressions?
     
  6. Jul 10, 2016 #5

    cnh1995

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    Are you familiar with the componendo-dividendo property of equal ratios?
    If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d).
     
    Last edited: Jul 10, 2016
  7. Jul 10, 2016 #6
    I really would like to thank you guys I got it. I was hesitating to post the question as I thought it was too obvious. Thanks again :smile:
     
  8. Jul 11, 2016 #7

    James R

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    Gold Member

    In case somebody else reading this didn't get the answer, we can take an expression like this:

    ##\sin \theta - \cos \theta \le \mu (\cos \theta + \sin \theta)##

    and divide both sides by ##\cos \theta## to get

    ## \tan \theta - 1 \le \mu (1 + \tan \theta)##

    then go from there.
     
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