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anonymous

  • 5 years ago

Which of the following is equivalent to the expression below? 1+sec(theta) ----------------- tan(theta)+sin(theta)

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  1. anonymous
    • 5 years ago
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    \[\frac{1 + sec(\theta)}{tan(\theta) + sin(\theta)} = \frac{1 + \frac{1}{cos(\theta)}}{\frac{sin(\theta)}{cos(\theta)} + sin(\theta)}\]

  2. anonymous
    • 5 years ago
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    I dunno what options you have, these trig identities can get complicated.

  3. anonymous
    • 5 years ago
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    csc(theta) tan(theta) -1 sec(theta) cot(theta)

  4. anonymous
    • 5 years ago
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    Well, from the denominator you can factor out a sin(theta) from each term.. what do you have then?

  5. anonymous
    • 5 years ago
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    uh (1+1/cos(theta))/(cos(theta))?

  6. anonymous
    • 5 years ago
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    no. try factoring again \[\frac{sin\theta}{cos\theta} + sin\theta = ?\]

  7. anonymous
    • 5 years ago
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    sec(theta)

  8. anonymous
    • 5 years ago
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    that's the answer yes, but not what I was asking.. I assume you saw the solution after you factored it?

  9. anonymous
    • 5 years ago
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    err actually no, that's not the answer. nvm

  10. anonymous
    • 5 years ago
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    what do you get when you factor out a sin(theta) from that expression in the denominator?

  11. anonymous
    • 5 years ago
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    uhhhh ok how do I factor it out?

  12. anonymous
    • 5 years ago
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    How do you factor out the a from 2a + ba ?

  13. anonymous
    • 5 years ago
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    Or factor out a 2 from (4 + 6a)

  14. anonymous
    • 5 years ago
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    sin(theta)(1/cos(theta)

  15. anonymous
    • 5 years ago
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    close

  16. anonymous
    • 5 years ago
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    try this one: factor completely (10x+2)

  17. anonymous
    • 5 years ago
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    2(5x+1)

  18. anonymous
    • 5 years ago
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    right. Now try \(\frac{a}{b} + a\)

  19. anonymous
    • 5 years ago
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    or we can write it as\[\frac{1}{b}*a + a\]

  20. anonymous
    • 5 years ago
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    if that makes it clearer

  21. anonymous
    • 5 years ago
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    sin(θ)(1+cos(θ))/cos(θ)

  22. anonymous
    • 5 years ago
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    no.

  23. anonymous
    • 5 years ago
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    \[\frac{a}{b} + a = a(\frac{1}{b} + 1). a = sin(\theta), b = cos(\theta) \]\[\implies \frac{sin(\theta)}{cos(\theta)} + sin(\theta) = sin\theta(\frac{1}{cos\theta} + 1)\]

  24. anonymous
    • 5 years ago
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    oh ok.... i guess i was kinda close, but not really

  25. anonymous
    • 5 years ago
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    so from there you should see something nice to cancel and get a simple answer

  26. anonymous
    • 5 years ago
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    what like the 1?

  27. anonymous
    • 5 years ago
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    no

  28. anonymous
    • 5 years ago
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    Like the [1 + 1/cos(theta)] you have on top and bottom

  29. anonymous
    • 5 years ago
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    what equation are you looking at?

  30. anonymous
    • 5 years ago
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    \[\frac{1+\frac{1}{cos\theta}}{sin\theta (1 + \frac{1}{cos\theta})}\]

  31. anonymous
    • 5 years ago
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    so all I have left is 1/sin(theta)?

  32. anonymous
    • 5 years ago
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    indeed

  33. anonymous
    • 5 years ago
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    which equals...csc(theta)

  34. anonymous
    • 5 years ago
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    indeed

  35. anonymous
    • 5 years ago
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    wow sorry about making that so difficult thank you so much!

  36. anonymous
    • 5 years ago
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    lol, np

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