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pottersheep

  • 4 years ago

Please help me prove this trig identity.... (secx - cosecx)/(tanx - cotx) = (tanx - cotx) /(secx - cosecx)

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  1. myininaya
    • 4 years ago
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    i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions

  2. myininaya
    • 4 years ago
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    and then you might have to use some other identities

  3. moses_
    • 4 years ago
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    break up the LHS into \[\frac{sec x}{tan x - cot x} + \frac{-csc x}{tan x - cot x} \]

  4. pottersheep
    • 4 years ago
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    Oh....I'll try it that way then but Im not sure if I can reach the answer still.... And how can I change them to cos and sin? they are not squared :(

  5. moses_
    • 4 years ago
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    my advice would be to multiply by one using \[\frac{sinx}{sinx} \]

  6. moses_
    • 4 years ago
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    he means turn sec x into \[\frac{1}{cos x} \] etc...

  7. moses_
    • 4 years ago
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    for multiplying by one like i said above, dont use sin...use something that will reduce everything...

  8. pottersheep
    • 4 years ago
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    hmmm

  9. pottersheep
    • 4 years ago
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    i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too

  10. bluebrandon
    • 4 years ago
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    try taking everything in terms of cos and sin and then simplifying as much as you can. Don't leave any sec, csc, tan or cot

  11. Mertsj
    • 4 years ago
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    |dw:1322797409581:dw|

  12. myininaya
    • 4 years ago
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    i'm a her \[\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}\] now for both sides we want to clear the compound fraction action going on \[\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)} \] \[\frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}=\frac{\sin^2(x)-\cos^2(x)}{\sin(x)-\cos(x)}\] now we can some factoring on both sides \[\frac{\sin(x)-\cos(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}=\frac{(\sin(x)+\cos(x))(\sin(x)-\cos(x))}{\sin(x)-\cos(x)}\] so we have \[\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)\] ...

  13. Mertsj
    • 4 years ago
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    Now. Use the same technique on the right side.

  14. myininaya
    • 4 years ago
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    this doesn't look like an identity

  15. moses_
    • 4 years ago
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    oh my apologies

  16. pottersheep
    • 4 years ago
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    Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake

  17. myininaya
    • 4 years ago
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    it is not

  18. pottersheep
    • 4 years ago
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    oh wait a minute if I change both sides then I can get them to be equal

  19. myininaya
    • 4 years ago
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    try x=pi/3 both sides have different outputs

  20. myininaya
    • 4 years ago
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    one side is approximately 1.366 the other side is .73205

  21. Mertsj
    • 4 years ago
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    No. It is not. You have shown that they indeed are reciprocals which was apparent by a casual inspection of the original problem.

  22. myininaya
    • 4 years ago
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    the equation does hold for some values but not all which makes it not an identity

  23. pottersheep
    • 4 years ago
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    Thanks you guys =) i guess i need much more practice

  24. bluebrandon
    • 4 years ago
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    when I simplified both sides after making them in terms of sin and cos I got (sinx-cosx)(sinx^2-cosx^2) ------------------------ cosx^2*sinx^2 for both the left hand side and the right hand side

  25. bluebrandon
    • 4 years ago
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    start with|dw:1322798230411:dw|

  26. Mertsj
    • 4 years ago
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    So how did you get the denominator to be a product instead of a difference?

  27. myininaya
    • 4 years ago
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    maybe he combine the fractions on top and on bottom but cos(x)sin(x) in the denominator of the bottom fraction would cancel with cos(x)sin(x) with the top's denominator

  28. myininaya
    • 4 years ago
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    so you wouldn't have that product anymore

  29. pottersheep
    • 4 years ago
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    yeah

  30. pottersheep
    • 4 years ago
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    omg! this is the second day we started trig functions and our teacher barely taught us anything......i wish she would start off with simpler question....

  31. pottersheep
    • 4 years ago
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    i still have like 18 questions to go...sigh...

  32. pottersheep
    • 4 years ago
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    trig identities :P not functions

  33. myininaya
    • 4 years ago
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    anyways potter we have already shown that this is not an identity do you have any questions as to why we concluded that?

  34. Mertsj
    • 4 years ago
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    Also, potter, are you absolutely sure that you copied the problem correctly?

  35. pottersheep
    • 4 years ago
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    Nope I understand...thank you everyone :) now i need to practice it

  36. myininaya
    • 4 years ago
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    ok have fun

  37. pottersheep
    • 4 years ago
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    lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol

  38. myininaya
    • 4 years ago
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    is potter okay?

  39. pottersheep
    • 4 years ago
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    yep potter's fine :)

  40. myininaya
    • 4 years ago
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    harry potter? lol is that what you were thinking?

  41. pottersheep
    • 4 years ago
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    yeah xD

  42. myininaya
    • 4 years ago
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    it wouldn't be so bad to be the most powerful wizard of all time

  43. pottersheep
    • 4 years ago
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    yeah :( then i woulnt have to worry about these math problems lol

  44. myininaya
    • 4 years ago
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    you could just turn a math problem you didn't feel like working on into a bunny

  45. pottersheep
    • 4 years ago
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    LOL hahaha

  46. pottersheep
    • 4 years ago
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    too bad i cant give u another medal for that :P

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