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Homework Help: Trig Identity 3

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data
    (1-cos^2x)(1+tan^2x) = tan^2x


    2. Relevant equations
    N/A


    3. The attempt at a solution
    (1-cos^2x)(1+tan^2x) = tan^2x
    L.S.
    = (sin^2x)(1+sin^2x/cos^2x)
    = sin^2x+(sin^4x/cos^2x)

    Now, I get a common denominator, but it's not doing anything for me. Did I do the right thing in converting the tan?
     
    Last edited: Jan 8, 2008
  2. jcsd
  3. Jan 8, 2008 #2
    Mess with the left side ... what is 1-cos^(2)x and convert tan^(2)x into sine/cosine.
     
  4. Jan 8, 2008 #3
    Already did that on the first post. See, I missed all the lessons at school so I'm trying to piece everything in by myself. Can you please take a look at what I did and specify the next steps?
     
  5. Jan 8, 2008 #4
    You did not do this step correctly, do it again and you will get an identity that simplifies everything.
     
  6. Jan 8, 2008 #5
    What I'm getting is below.

    = sin^2x+(sin^4x/cos^2x)
    = ((sin^2x)(cos^x)+sin^4x)/cos^2x

    The cos^2x is the common denominator.
     
  7. Jan 8, 2008 #6
    [tex]\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}[/tex]

    [tex]\sin^{2}x\left(1+\frac{\sin^{2}x}{\cos^{2}}\right)=\tan^{2}x[/tex]

    [tex]\sin^{2}x\left(1\times\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}}\right)=\tan^{2}x[/tex]
     
  8. Jan 8, 2008 #7
    Alright that clears up a lot. So I can make a 1 into cos/cos or sin/sin. Makes things a lot easier. Thank you.
     
  9. Jan 9, 2008 #8
    Tornaer:

    there is an identity relating to sec^x and 1+tan^2x. Find it out and you'll get your answer in 2 lines or less.
     
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