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Proving This Trigonometric Identity

  1. May 2, 2013 #1
    1. Prove:


    [tex]\frac{cscx +cotx}{cscx-cotx} = \frac{1+2cosx+cos^2x}{sin^2x}[/tex]






    2. Relevant equations
    tanx = [itex]\frac{sinx}{cosx}[/itex]
    cotx = [itex]\frac{cosx}{sinx}[/itex]
    cscx
    secx
    cotx
    [itex]sin^2x + cos^2x = 1[/itex]




    3. The attempt at a solution

    Left side:
    =[itex]\frac{cscx +cotx}{cscx-cotx}[/itex]
    =[itex]\frac{1/sinx + cosx / sinx}{1/sinx - cosx/sinx}[/itex]
    =[itex]\frac{1+cosx}{1-cosx}[/itex]
     
    Last edited by a moderator: May 2, 2013
  2. jcsd
  3. May 2, 2013 #2
    You are almost there! What could you multiply the numerator and denominator of your fraction to get the RHS?
     
  4. May 2, 2013 #3
    You are almost there. What is sin^2x in terms of cos^2x?

    EDIT: Just a minute late. -.-
     
  5. May 2, 2013 #4
    =[itex]\frac{1+cosx}{1-cosx}[/itex]
    =[itex]\frac{1+cosx}{1-cosx}[/itex] × [itex]\frac{1+cosx}{1+cosx}[/itex]
    = [itex]\frac{1+2cosx + cos^2x}{1-cos^2x}[/itex]
    =[itex]\frac{1+2cosx+cos^2x}{sin^2x}[/itex]

    thanks guys
     
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