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anonymous

  • one year ago

Which of the following world be an acceptable first step in simplifying the expression tanx/(1+secx)?

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  1. anonymous
    • one year ago
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  2. jim_thompson5910
    • one year ago
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    your thoughts?

  3. anonymous
    • one year ago
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    I read through the notes like 10 times and still don't understand

  4. jim_thompson5910
    • one year ago
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    |dw:1440198581765:dw|

  5. jim_thompson5910
    • one year ago
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    |dw:1440198631242:dw|

  6. anonymous
    • one year ago
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    isn't it eliminating the denominator ?

  7. jim_thompson5910
    • one year ago
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    hint: think of the difference of squares formula

  8. anonymous
    • one year ago
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    the (b/2)^2 ?

  9. jim_thompson5910
    • one year ago
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    You've seen this formula before hopefully a^2 - b^2 = (a-b)(a+b)

  10. anonymous
    • one year ago
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    oh yeah. I'm used to the (a-b)2 but I know what you're talking about

  11. anonymous
    • one year ago
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    (a-b)^2 *

  12. jim_thompson5910
    • one year ago
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    so the idea is that we multiply top and bottom by 1-sec(x), then we get this |dw:1440198998954:dw|

  13. jim_thompson5910
    • one year ago
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    |dw:1440199030582:dw|

  14. jim_thompson5910
    • one year ago
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    |dw:1440199066043:dw|

  15. jim_thompson5910
    • one year ago
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    |dw:1440199086975:dw|

  16. jim_thompson5910
    • one year ago
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    what is 1-sec^2 equal to?

  17. anonymous
    • one year ago
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    umm , the first number to pop up in my head was 1 lol

  18. jim_thompson5910
    • one year ago
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    nope

  19. jim_thompson5910
    • one year ago
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    look at this identity sheet http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

  20. jim_thompson5910
    • one year ago
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    look on page 2 where it says "Pythagorean Identities"

  21. anonymous
    • one year ago
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    tan^2 theta ?

  22. jim_thompson5910
    • one year ago
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    yep

  23. jim_thompson5910
    • one year ago
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    |dw:1440199398015:dw|

  24. jim_thompson5910
    • one year ago
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    |dw:1440199419561:dw|

  25. jim_thompson5910
    • one year ago
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    so the original expression is equivalent to \[\Large \frac{1-\sec(x)}{\tan(x)}\]

  26. anonymous
    • one year ago
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    okay, so the answer would be a ?

  27. jim_thompson5910
    • one year ago
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    correct

  28. anonymous
    • one year ago
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    thank you!

  29. jim_thompson5910
    • one year ago
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    no problem

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