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anonymous

  • one year ago

Trig / Pre Cal/ identities Guide me please. \( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1 \)

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  1. anonymous
    • one year ago
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    It is suppose to be. \( \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

  2. anonymous
    • one year ago
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    \( \large \rm \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x} \)

  3. anonymous
    • one year ago
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    These things are killing me lol

  4. anonymous
    • one year ago
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    not an identity

  5. xapproachesinfinity
    • one year ago
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    okay so we have \[\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\]

  6. anonymous
    • one year ago
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    I got that they're reciprocals

  7. xapproachesinfinity
    • one year ago
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    if so we can do this \[\Huge \frac{\tan x}{\sec x+1}=\frac{\sec x+1}{\tan x}\] clearly this is not truee

  8. xapproachesinfinity
    • one year ago
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    try some vlues if you have doubt

  9. anonymous
    • one year ago
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    Yes but I am trying to get from the left side to the right. \(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }\) Do I times by cos?? \(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1} \times cos x\)

  10. anonymous
    • one year ago
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    The identity is false is what he's saying. You won't be able to get from the left to the right.

  11. xapproachesinfinity
    • one year ago
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    now it does not work! left does not equal right

  12. xapproachesinfinity
    • one year ago
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    plug in some values to be sure if you have doubt

  13. xapproachesinfinity
    • one year ago
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    no* for (now)

  14. anonymous
    • one year ago
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    The original problem is \(\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x }\)

  15. xapproachesinfinity
    • one year ago
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    0 for example is not good left give you 0 right gives you undefined 2/0 if it is an identity it should not be that way

  16. anonymous
    • one year ago
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    I did the left side assuming that is what we have to do.

  17. xapproachesinfinity
    • one year ago
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    is the left 1/cosx -1 or 1/cosx +1 you just changed in your last reply?

  18. xapproachesinfinity
    • one year ago
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    in that case it is an identity

  19. xapproachesinfinity
    • one year ago
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    see that you give us the wrong thing from the start

  20. xapproachesinfinity
    • one year ago
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    if it is a minus on the bottom it is an identity since tan^2x=sec^2x-1

  21. anonymous
    • one year ago
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    This is the original problem. \(\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x } \) How do we know to work the numerator or the denominator or both?

  22. anonymous
    • one year ago
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    For instance, should I \(\Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x } \) OR \(\Huge \frac{\frac{sin}{cos}}{{\sec x}-1 }=\frac{\sec x+1}{\tan x } \) OR \(\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1} \)

  23. xapproachesinfinity
    • one year ago
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    usually when you prove something you should get from left to right of vice versa so taking left \[\frac{\tan x}{\sec x=1}=\frac{\tan x(\sec x+1)}{(\sec x-1)(\sec x+1)}\] \[=\frac{\tan x(\sec x+1)}{\tan^2x }=\frac{\sec x+1}{\tan x} ~~~Q.E.D\]

  24. xapproachesinfinity
    • one year ago
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    you didn't need to do all that just leave tan and sec that way

  25. xapproachesinfinity
    • one year ago
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    you are making it hard

  26. anonymous
    • one year ago
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    There are no rules though

  27. xapproachesinfinity
    • one year ago
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    what rules?

  28. xapproachesinfinity
    • one year ago
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    did you get what i did?

  29. anonymous
    • one year ago
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    No

  30. xapproachesinfinity
    • one year ago
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    if we do it your way we have \[\frac{\sin x}{1-\cos x}=\frac{\sin x(1+\cos x)}{1-\cos^2x}=\frac{1+\cos x}{\sin x}\] multiplyinh by 1/cos top and bottom we get \[=\frac{\sec x+1}{\tan x}\]

  31. xapproachesinfinity
    • one year ago
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    perhaps i'm going fast?

  32. xapproachesinfinity
    • one year ago
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    i skip some stuff given that you know how to deal with algebra

  33. anonymous
    • one year ago
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    I don't get it. How do I know to work the numerator or the denominator or both. How do I know what method to use when there are no rules to guy you.

  34. anonymous
    • one year ago
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    For instance, how did you do it your way?

  35. xapproachesinfinity
    • one year ago
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    like \[\huge \frac{\frac{a}{b}}{\frac{1}{b}-1}=\frac{a}{b(\frac{1}{b}-1)}=\frac{a}{1-b}\]

  36. xapproachesinfinity
    • one year ago
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    this is what i did with your way changing tan to sin/cos and sec to 1/cos

  37. anonymous
    • one year ago
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    Yes that is my way.

  38. anonymous
    • one year ago
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    But there are other ways and some ways don't work

  39. xapproachesinfinity
    • one year ago
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    yes once you get sin x/1-cosx you multiply by 1+cosx top and bottom

  40. anonymous
    • one year ago
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    Correct

  41. xapproachesinfinity
    • one year ago
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    what other Ways do you mean if you are using identities they should work

  42. xapproachesinfinity
    • one year ago
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    my way is no different than yours i still had to multiply top and bottom be 1+sec x to get sec^2x-1 on the bottom which is the same as tan^2x

  43. xapproachesinfinity
    • one year ago
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    this is just a matter of knowing your identities and how to use them

  44. anonymous
    • one year ago
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    For instance \( \Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x } \) \( \Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x } \)

  45. xapproachesinfinity
    • one year ago
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    it is still working since top becomes sinx

  46. xapproachesinfinity
    • one year ago
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    gotta go! i think you got this now?

  47. anonymous
    • one year ago
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    Ok so we have \(\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x } \) \(\Huge \frac{sin x}{1-cos x}=\frac{\sec x+1}{\tan x } \)

  48. xapproachesinfinity
    • one year ago
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    yes!

  49. xapproachesinfinity
    • one year ago
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    i explained above how you obtain the right side

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