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Proving Trigonometric Identites

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    tan x + 1/tan x - 1 = sec x + csc x/sec x - csc x

    2. Relevant equations

    3. The attempt at a solution

    Ive tried working with the right side and have gotten as far as sin x/cosxsinx - cos x/cosxsinx

    Im not sure if im just out to lunch, or on the right track. Any help would be awesome!

  2. jcsd
  3. Jan 13, 2009 #2


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    Homework Helper
    Gold Member

    Your expression is ambiguous the way you have written it. (Use either brackets or LaTeX)

    Do you mean [tex]\frac{\tan(x)+1}{\tan(x)-1}=\frac{\sec(x)+\csc(x)}{\sec(x)-\csc(x)}[/tex] ???

    If so, try working with the LHS: Express everything in terms of sines and cosines and then divide both the numerator and denominator by [itex]\sin(x)[/itex]
  4. Jan 14, 2009 #3
    im not getting it. should i turn the 1 into sines or cosines? Or even how do I know to do that?
  5. Jan 14, 2009 #4
    No you should turn the tan into sin/cos and the find common denominators, cancel, etc. etc.
  6. Jan 14, 2009 #5
    Oh I did that too.

    I had sin/cos + cos/cos all divided by sin/cos - cos/cos
  7. Jan 14, 2009 #6
    Well a/c + b/c = (a+b)/c and also (a/b)/(c/d) = (ad)/(bc) I heard maybe you should use that.
  8. Jan 14, 2009 #7
    Well I heard that gives me sin x cos x + cos^2 x / sin x cos x - cos ^2 x
  9. Jan 14, 2009 #8
    Umm why haven't you cancelled the cos
  10. Jan 14, 2009 #9
    Your right, now I have sin x + cos x / sin x - cos x ?
  11. Jan 14, 2009 #10
    Now compare that to what you are trying to get to...
  12. Jan 14, 2009 #11
    I got it. I went with the RHS side though, and converted it to the LHS. Thanks for your help!
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