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Homework Help: Trig help

  1. May 22, 2005 #1
    i need help showing the folowing is valid:

    tan^4(w) + 2tan^2(w) +1 = sec^4(w)

    I am pretty lost, I know I should expand the left side and go from there. SO far I have gotten:

    [sin^4(w)/cos^4(w)] + [2sin^2(w)/cos^2(w)] + [sin^2(w) +cos^2(w)]

    and then I am stuck when I try and add these terms together, Imay have made mistakes some where along the line in adding the fractions, could someone point me in the right direction? am I right so far?
    the farthest i have gotten on the left side is:
    [sin^4(w) + 2sin^2(w)cos^2(w) + sin^2(w)cos^4(w) +cos^6(w)] /cos^4(w)
     
    Last edited: May 22, 2005
  2. jcsd
  3. May 22, 2005 #2

    Curious3141

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    Homework Helper

    Observe that the LHS is just [tex]{(1 + \tan^2{w})}^2 = {(\sec^2{w})}^2 = \sec^4{w}[/tex]

    If you're not allowed to assume that [tex]1 + \tan^2{w} = \sec^2{w}[/tex], just divide [tex]\sin^2{w} + \cos^2{w} = 1[/tex] throughout by [tex]\cos^2{w}[/tex] and see what you get.
     
  4. May 23, 2005 #3
    Another way out if you're not "allowed" to do some other things is to subtract 1 from each side of the equality to be proved and use a variant of the identity given by Curious3141.

    Note that what you have to prove is not strictly identity for all [itex]w[/itex], unless you bend the definition of an identity to accomodate it. The terms on the left hand side are not defined when [itex]x = (2n+1)\frac{\pi}{2}[/itex] where n is an integer. Same goes for the term on the right hand side. However, it would be better to say that the terms tend to [itex]\infty[/itex] as x approaches this value from the left or right and hence, this minor argument does not really matter much.

    Cheers
    Vivek
     
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