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abichu

  • 4 years ago

6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1

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  1. myininaya
    • 4 years ago
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    we want to show the second part?

  2. abichu
    • 4 years ago
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    yea!

  3. myininaya
    • 4 years ago
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    what is T?

  4. abichu
    • 4 years ago
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    PT= PROVE THAT

  5. myininaya
    • 4 years ago
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    so we get to suppose \[\sec(\theta)+\tan(\theta)=P\] and we want to show \[PT \sin(\theta)=\frac{P^2-1}{P^2+1}\]?

  6. myininaya
    • 4 years ago
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    oh ok

  7. abichu
    • 4 years ago
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    and guys can you answer my prev.Questions too?! please..

  8. myininaya
    • 4 years ago
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    \[\sec(\theta)+\tan(\theta)=\frac{1}{\cos(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}=\frac{1+\sin(\theta)}{\cos(\theta)}=P\] thinking....

  9. myininaya
    • 4 years ago
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    \[\frac{1+\sin(\theta)}{\cos(\theta)}*\frac{1-\sin(\theta)}{1-\sin(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}\] \[=\frac{\cos^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\frac{\cos(\theta)}{1-\sin(\theta)}=P\] hmmm....

  10. abichu
    • 4 years ago
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    solved?

  11. myininaya
    • 4 years ago
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    no lol

  12. myininaya
    • 4 years ago
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    i havent solved it yet

  13. myininaya
    • 4 years ago
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    maybe i should try this on paper first

  14. abichu
    • 4 years ago
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    what bout my prev.sums? can you solve it?

  15. myininaya
    • 4 years ago
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    let me try this one first

  16. abichu
    • 4 years ago
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    sure! :)

  17. myininaya
    • 4 years ago
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    if \[\sec(\theta)+\tan(\theta)=P\] then \[(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=P(\sec(\theta)-\tan(\theta))\] but \[(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=\sec^2(\theta)-\tan^2(\theta)=1\] so we have \[1=P(\sec(\theta)-\tan(\theta))\] => \[\sec(\theta)-\tan(\theta)=\frac{1}{P}\] lets see if we can use this

  18. myininaya
    • 4 years ago
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    lets see we have that \[\sec(\theta)+\tan(\theta)=\frac{1+\sin(\theta)}{\cos(\theta)}\] and \[\sec(\theta)-\tan(\theta)=\frac{1-\sin(\theta)}{\cos(\theta)}\] adding these together we get \[\frac{1+\sin(\theta)}{\cos(\theta)}+\frac{1-\sin(\theta)}{\cos(\theta)}=P+\frac{1}{P}\] \[\frac{2}{\cos(\theta)}=\frac{P^2+1}{P}\] \[\frac{2P}{P^2+1}=\cos(\theta)\]

  19. myininaya
    • 4 years ago
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    but if we imagine we have a right triangle and we draw theta somewhere then cos tells adj/hyp=2p/(P^2+1) we can find the opposite of side of theta opp^2+adj^2=hyp^2 \[opp^2=hyp^2-adj^2=(P^2+1)^2-(2P)^2=P^4+2P^2+1-2^2P^2\] so we have \[opp=\sqrt{P^4-2P^2+1}\] but sin=opp/hyp

  20. myininaya
    • 4 years ago
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    \[\sin(\theta)=\frac{\sqrt{P^4-2P^2+1}}{P^2+1}=\frac{\sqrt{(P^2-1)^2}}{P^2+1}\]

  21. myininaya
    • 4 years ago
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    \[\sin(\theta)=\frac{P^2-1}{P^2+1}\]

  22. myininaya
    • 4 years ago
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    any questions?

  23. abichu
    • 4 years ago
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    yea..can you solve the other problems?

  24. myininaya
    • 4 years ago
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    zarkon?

  25. abichu
    • 4 years ago
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    what?

  26. myininaya
    • 4 years ago
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    i was looking for someone

  27. Zarkon
    • 4 years ago
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    yes

  28. abichu
    • 4 years ago
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    and can you temme one thing..can you temme the stepin correct order so that i can copy it in my rough boook! ;)

  29. myininaya
    • 4 years ago
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    hey zarkon i had to think on it isn't that crazy?

  30. Zarkon
    • 4 years ago
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    yes

  31. myininaya
    • 4 years ago
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    the problem

  32. Zarkon
    • 4 years ago
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    trying to think if there is an easier way

  33. myininaya
    • 4 years ago
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    i have no clue it was hard for me to see that way

  34. myininaya
    • 4 years ago
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    what is temme?

  35. myininaya
    • 4 years ago
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    i don't know that word

  36. myininaya
    • 4 years ago
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    use small words please lol

  37. myininaya
    • 4 years ago
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    well i mean i guess its small but i don't know it

  38. myininaya
    • 4 years ago
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    tell me?

  39. myininaya
    • 4 years ago
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    temme=tell me?

  40. myininaya
    • 4 years ago
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    i thought i showed u the steps zarkon thinks there may be an easier way though

  41. myininaya
    • 4 years ago
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    zarkon i did this for you :)

  42. myininaya
    • 4 years ago
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    lol

  43. Zarkon
    • 4 years ago
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    ;)

  44. Zarkon
    • 4 years ago
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    how about this...

  45. myininaya
    • 4 years ago
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    and for abichu dont feel left out

  46. Zarkon
    • 4 years ago
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    get to \[\frac{1+sin(\theta)}{\cos(\theta)}=p\] ...

  47. abichu
    • 4 years ago
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    i am soo weak in math..that's why am seeking help from you guys! if you dont mind help me! ;)

  48. myininaya
    • 4 years ago
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    we are helping

  49. Zarkon
    • 4 years ago
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    \[1+\sin(\theta)=p\cos(\theta)\]

  50. myininaya
    • 4 years ago
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    i showed you one way and zarkon is gonna show you another it appears

  51. Zarkon
    • 4 years ago
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    \[1+\sin(\theta)=p\sqrt{1-\sin^2(\theta)}\]

  52. Zarkon
    • 4 years ago
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    square both sides...

  53. myininaya
    • 4 years ago
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    omg i see where you are going

  54. Zarkon
    • 4 years ago
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    \[1+2\sin(\theta)+\sin^2(\theta)=p^2(1-\sin^2(\theta))\]

  55. myininaya
    • 4 years ago
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    you going to write is as a quadratic

  56. myininaya
    • 4 years ago
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    and solve for sin(theta)

  57. abichu
    • 4 years ago
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    can you tel me from the 1st step if you dot mind?!

  58. Zarkon
    • 4 years ago
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    move stuff to one side \[1-p^2+2\sin(\theta)+(1+p^2)\sin^2(\theta)=0\]

  59. myininaya
    • 4 years ago
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    brillant!

  60. Zarkon
    • 4 years ago
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    \[1-p^2+2x+(1+p^2)x^2=0\]

  61. Zarkon
    • 4 years ago
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    solve this quadratic ;)

  62. Zarkon
    • 4 years ago
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    fun stuff :)

  63. myininaya
    • 4 years ago
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    yes

  64. myininaya
    • 4 years ago
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    you are a showoff lol

  65. Zarkon
    • 4 years ago
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    lol...now abichu has two ways to do it

  66. abichu
    • 4 years ago
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    i donno how to solve it quadratic..can you doo it n give?

  67. myininaya
    • 4 years ago
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    \[x=\frac{-2 \pm \sqrt{(2)^2-4(1+p^2)(1-p^2)}}{2(1+p^2)}\]

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