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pottersheep Group Title

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

  • 2 years ago
  • 2 years ago

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  1. myininaya Group Title
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    i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions

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

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

    • 2 years ago
  4. pottersheep Group Title
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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 :(

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

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

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

    • 2 years ago
  8. pottersheep Group Title
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    hmmm

    • 2 years ago
  9. pottersheep Group Title
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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

    • 2 years ago
  10. bluebrandon Group Title
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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

    • 2 years ago
  11. Mertsj Group Title
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    |dw:1322797409581:dw|

    • 2 years ago
  12. myininaya Group Title
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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 \[\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)\] ...

    • 2 years ago
  13. Mertsj Group Title
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    Now. Use the same technique on the right side.

    • 2 years ago
  14. myininaya Group Title
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    this doesn't look like an identity

    • 2 years ago
  15. moses_ Group Title
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    oh my apologies

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

    • 2 years ago
  17. myininaya Group Title
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    it is not

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

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

    • 2 years ago
  20. myininaya Group Title
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    one side is approximately 1.366 the other side is .73205

    • 2 years ago
  21. Mertsj Group Title
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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.

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

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

    • 2 years ago
  24. bluebrandon Group Title
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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

    • 2 years ago
  25. bluebrandon Group Title
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    start with|dw:1322798230411:dw|

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

    • 2 years ago
  27. myininaya Group Title
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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

    • 2 years ago
  28. myininaya Group Title
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    so you wouldn't have that product anymore

    • 2 years ago
  29. pottersheep Group Title
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    yeah

    • 2 years ago
  30. pottersheep Group Title
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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....

    • 2 years ago
  31. pottersheep Group Title
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    i still have like 18 questions to go...sigh...

    • 2 years ago
  32. pottersheep Group Title
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    trig identities :P not functions

    • 2 years ago
  33. myininaya Group Title
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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?

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

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

    • 2 years ago
  36. myininaya Group Title
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    ok have fun

    • 2 years ago
  37. pottersheep Group Title
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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

    • 2 years ago
  38. myininaya Group Title
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    is potter okay?

    • 2 years ago
  39. pottersheep Group Title
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    yep potter's fine :)

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

    • 2 years ago
  41. pottersheep Group Title
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    yeah xD

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

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

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

    • 2 years ago
  45. pottersheep Group Title
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    LOL hahaha

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

    • 2 years ago
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