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Hannah_Ahn

  • 2 years ago

prove the identity csc theta over tan theta + cot theta = cos theta

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  1. Zed
    • 2 years ago
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    \[\frac{ \frac{1}{sin {\theta}}}{tan{\theta}}+\frac{1}{tan{\theta}}=cos{\theta}\]

  2. Hannah_Ahn
    • 2 years ago
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    I am sorry, how did that happen?

  3. Zed
    • 2 years ago
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    Sorry I will write it up first and post it with steps. Just a moment

  4. Hannah_Ahn
    • 2 years ago
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    Sweet, thanks!

  5. Yifan12879
    • 2 years ago
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    does it mean that (1/sin x)^2+(1/sin x)=1?

  6. Yifan12879
    • 2 years ago
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    if you divide the cos theta from both sides

  7. Zed
    • 2 years ago
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    Yes correct

  8. Hannah_Ahn
    • 2 years ago
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    I still would appreciate to see the steps plz :s

  9. Yifan12879
    • 2 years ago
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    but why? let x=pi/6 then 4+2=1??

  10. Zed
    • 2 years ago
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    yes I'm having that issue too. Are you sure it's written down right?

  11. Hannah_Ahn
    • 2 years ago
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    \[\csc \theta \over \tan \theta +\cot \theta\] = [\cos \theta\]

  12. Zed
    • 2 years ago
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    I see now, next time put the tan theta + cot theta in brackets to make it clearer.

  13. Yifan12879
    • 2 years ago
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    multiply the LHS by sinx cosx

  14. Hannah_Ahn
    • 2 years ago
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    gotcha. sorry about that.

  15. Yifan12879
    • 2 years ago
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    i mean csc x /(tan x + cot x)= cscx sinx cosx / (tanx sinx cosx+cotx sinx cosx)=cosx/(sin^2 x+cos^2 x)=cosx

  16. Yifan12879
    • 2 years ago
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    im sorry it should be theta...

  17. Hannah_Ahn
    • 2 years ago
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    why did you multiply each sides by sinx cosx? how could u tell?

  18. Zed
    • 2 years ago
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    \[\frac{1/sin{\theta}}{tan{\theta}+1/tan{\theta}}=cos{\theta}\]\[1/sin{\theta}=tan{\theta}cos{\theta}+cos{\theta}/tan{\theta}\]\[1/sin{\theta}=sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=sin^2{\theta}/sin{\theta}+cos^2{\theta}/sin{\theta}\]\[1/sin{\theta}=1/sin{\theta}\]\[1=1\]

  19. Zed
    • 2 years ago
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    Here's a different way to go about it :)

  20. Yifan12879
    • 2 years ago
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    because there are sinx cosx in the denominator

  21. Hannah_Ahn
    • 2 years ago
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    my teacher doesn't let me combite the left side and the right side.. in this case i would need to make the left side cos theta. :S

  22. Hannah_Ahn
    • 2 years ago
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    zed; yur way is something to think about, and it's cool. yet i won't b able to use your method

  23. Yifan12879
    • 2 years ago
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    @Zed you shouldnt give out a prove like this, you should write LHS=...=....=...=RHS

  24. Hannah_Ahn
    • 2 years ago
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    yifian; i don't really get how you got there.. :(

  25. Hannah_Ahn
    • 2 years ago
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    i can't figure it out

  26. Yifan12879
    • 2 years ago
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    which step?

  27. Hannah_Ahn
    • 2 years ago
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    is there other way to do it?

  28. Zed
    • 2 years ago
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    Thanks Yifan, I haven't done proof in a long time. \[LHS = \frac{1/sinx}{sinx/cosx+cosx/sinx}\] \[=\frac{1/sinx}{(sin^2x+cos^2x)/sinxcosx}\]\[=\frac{1/sinx}{1/sinxcosx}\]\[=\frac{1}{cosx}\]\[=RHS\]

  29. Zed
    • 2 years ago
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    sorry that second last line was supposed to be cosx

  30. Yifan12879
    • 2 years ago
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    yeah that is a good prove except that you substitute theta for x..

  31. Zed
    • 2 years ago
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    yeah i got tired of writing theta out

  32. Zed
    • 2 years ago
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    thanks Yifan

  33. Hannah_Ahn
    • 2 years ago
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    1 over cos x isn't same as cos x

  34. Zed
    • 2 years ago
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    It was a typo

  35. Hannah_Ahn
    • 2 years ago
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    i got that too.

  36. Hannah_Ahn
    • 2 years ago
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    well thanks for your time :)

  37. Hannah_Ahn
    • 2 years ago
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    i still don't get it . for some reason, im very confused. i will ask my teacher and I will let you know if there is a simpler way to do it if you are interested :P g'nite.

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