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Homework Help: Expression eqivalent to 1+sec/cos ? Explain

  1. Feb 3, 2005 #1
    Expression eqivalent to 1+sec/cos ?? Explain plz

    How do u find an expression equivalent to [tex] \frac {1+\sec\theta} {\cos\theta} [/tex] to be [tex] \frac {\cos\theta+1} {\cos^2\theta} [/tex]

    When I did this question I got my answer to be [tex] \frac {2} {\cos\theta}[/tex]

    Im not very good with identities can someone please explain to me how to do this question and other like it.
     
    Last edited: Feb 3, 2005
  2. jcsd
  3. Feb 3, 2005 #2

    learningphysics

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    Can you show your steps to getting that? I'll point out where you went wrong.
     
  4. Feb 3, 2005 #3
    [tex] \frac {sin^2\theta + \cos^2\theta} {\cos \theta} + \frac {\sin^2\theta+\cos^2\theta} {\cos\theta} (this is over \cos \theta) [/tex]
     
  5. Feb 3, 2005 #4

    learningphysics

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    Forget about [tex] sin^2\theta + \cos^2\theta =1 [/tex].

    Do the problem again, but this time only use:[tex]sec\theta=\frac{1}{cos\theta}[/tex] and try to simplify it.
     
    Last edited: Feb 3, 2005
  6. Feb 3, 2005 #5
    I have gone through so many of these types of questions but I have no idea of how to simplify it, I've looked everywhere for a similar example but just cant find anything, I dont know what to do... :cry: How do u solve these questions that ask for equivalent identities? :redface:
     
  7. Feb 3, 2005 #6

    cepheid

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    Well, did you take learningphysics' advice? Substitute 1/sec(x) every time you see cos(x) to get the expression in terms of secant, instead of cosine. That is definitely a step in the right direction.

    BTW, check back to your thread "plz explain". I posted the solution to your previous problem there, since you were very close to understanding it anyway.
     
  8. Feb 3, 2005 #7

    dextercioby

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    Use:
    [tex] \sec x =\frac{1}{\cos x} [/tex]

    and
    [tex] \frac{1+\frac{1}{a}}{b}=\frac{a+1}{ab} [/tex]

    which is an elementary property of fractions.

    Daniel.
     
  9. Feb 3, 2005 #8
    Can u explain how u got this elementary property of fractions, this might be the step I dont remember lol
     
  10. Feb 3, 2005 #9

    learningphysics

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    I plugged in [tex]sec\theta=\frac{1}{cos\theta}[/tex]

    I get:

    [tex] \frac {1+\frac{1}{cos\theta}} {\cos\theta} [/tex]

    Do something to the numerator and denominator so that you get a cleaner looking expression. Try different things... you can try adding the two terms in the numerator together (get a common denominator etc...)
     
  11. Feb 3, 2005 #10
    This is exactly where I am stuck!I tried multiplying the numerator and denominator by cos theta, but that didnt work, I took ur advice and im half way I made the numerators have a common denominator

    so I got [tex] 1+\frac{1}{\cos\theta} = \frac {\cos\theta} {\cos\theta}+ \frac {1}{\cos\theta} [/tex] This is only the numerator of the question my answer was [tex] \frac {\cos\theta+1} {\cos\theta} [/tex] But this is all over [tex] \cos\theta [/tex] what do u do next? How do you get cos^2theta in the denominator of the answer :yuck: ?
     
  12. Feb 3, 2005 #11

    learningphysics

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    Try working on the expression dextercioby gave... how can you rewrite the expression... try experimenting...

    That 1/a is messy, how can you rewrite the expression so that the 1/a is no longer there?
     
  13. Feb 3, 2005 #12

    dextercioby

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    [tex] \frac{1+\frac{1}{a}}{b}=\frac{\frac{a+1}{a}}{b}=\frac{a+1}{ab} [/tex]

    Daniel.
     
  14. Feb 4, 2005 #13

    learningphysics

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    Ok. I see where you are stuck now. The whole fraction is:
    [tex] \frac { \frac {\cos\theta} {\cos\theta}+ \frac {1}{\cos\theta}}{\cos\theta} [/tex]

    I'll rewrite the numerator, and the fraction becomes:
    [tex] \frac { \frac {\cos\theta +1}{\cos\theta}}{cos\theta} [/tex]

    Now the denominator of this fraction is [tex]\cos\theta[/tex]

    Instead of dividing by [tex]\cos\theta[/tex], I'm going to multiply by:
    [tex]\frac{1}{\cos\theta}[/tex]

    So we have:
    [tex]\frac {\cos\theta +1}{\cos\theta} * \frac{1}{\cos\theta}[/tex]

    Multiply the numerators, and multiply the denominators and we get:
    [tex] \frac {\cos\theta+1}{\cos^2\theta} [/tex]
     
  15. Feb 4, 2005 #14

    learningphysics

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    [tex]\frac{(\frac{a}{b})}{c} = \frac{a}{bc}[/tex]
     
    Last edited: Feb 4, 2005
  16. Feb 4, 2005 #15
    Why not just multiple top and bottom by cos....
    [tex] \frac {1+\sec\theta} {\cos\theta} [/tex]
    [tex] \frac {1+\sec\theta} {\cos\theta}(\frac{\cos\theta}{\cos\theta}) [/tex]
    [tex] \frac {\cos\theta+1} {\cos^2\theta} [/tex]

    because cos*sec =1 = cos*(1/cos) = 1
     
  17. Feb 4, 2005 #16
    :eek: WoW I think I get it thanks sooo much everyone, once again u all saved me!
     
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