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anonymous

  • one year ago

Can someone check my answer, I feel like I did it completely wrong!! 13. Verify the identity. Justify each step. sec (0) sec(0) ------------- - ------------- = 2csc * (0) csc(0)-cot(0) csc(0)+cot(0) (sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0))) --------------------------------------------------- = 2csc*(0) (csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0)) (sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0)) ------------------------------------------------------------ (csc^2(0)-cot^2(0)) =2csc*(0) (2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0))

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    sec (0) sec(0) ------------- - ------------- = 2csc * (0) csc(0)-cot(0) csc(0)+cot(0)

  3. anonymous
    • one year ago
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    (sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0))) --------------------------------------------------- = 2csc*(0) (csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0))

  4. anonymous
    • one year ago
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    (sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0)) ------------------------------------------------------------ (csc^2(0)-cot^2(0)) =2csc*(0)

  5. anonymous
    • one year ago
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    (2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0)) --------------------------------------------- = 2csc*(0) (csc^2(0)-cot^2(0))

  6. anonymous
    • one year ago
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    csc^2(0)-cot^2(0)=1 ((2csc*(2theta)+csc(0))-2csc*(2theta)+csc(0)) 2csc(0)=2csc(0)

  7. Nnesha
    • one year ago
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    confusion and confusion o,o how did you get 2 there http://prntscr.com/8qb8cf ?

  8. anonymous
    • one year ago
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    im not quite sure. I watched a video about how to do it but i feel like i didnt learn a thing lol could you help me

  9. Nnesha
    • one year ago
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    hmm sure i'll just start from http://prntscr.com/8qb8um not gonna start over by different way

  10. anonymous
    • one year ago
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    omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her

  11. Nnesha
    • one year ago
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    distribute \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] ignore theta for a sec

  12. Nnesha
    • one year ago
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    what's the definition of sec? like csc = 1/sin so sec = ?

  13. anonymous
    • one year ago
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    im not sure:/

  14. Nnesha
    • one year ago
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    oh okay thats fine but these are very important \[\sin \theta= \frac{1}{\csc \theta} ~~~\cos \theta =\frac{1}{\sec \theta }~~~ \tan \theta= \frac{1}{\cot \theta }\] sin cos tan and csc sec cot are reciprocal

  15. Nnesha
    • one year ago
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    \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] first look at the numerator combine like terms !let me know what u get :=)

  16. Nnesha
    • one year ago
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    like sec*csc and -sec*csc are like terms so sec*csc - sec*csc = ?

  17. anonymous
    • one year ago
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  18. Nnesha
    • one year ago
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    that is a different question right ?

  19. anonymous
    • one year ago
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    i redid the equation

  20. Nnesha
    • one year ago
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    i'm confused don't know how u got tan and cos at the beginning

  21. Nnesha
    • one year ago
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    the work u posted here is almost correct but there was a little mistake so i'll say keep the same method

  22. Nnesha
    • one year ago
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    \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] just two steps then we are done !

  23. anonymous
    • one year ago
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    oh shoot it was the wrong one im sorry. i guess ive been doing too much work today apparently lol ill just start from where you lfet off then

  24. Nnesha
    • one year ago
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    ohh i was so confused.

  25. anonymous
    • one year ago
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    my baddd :)

  26. Nnesha
    • one year ago
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    its okay \[\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }\] can you simplify the numerator just like simple algebra :=) combine like terms :=)

  27. anonymous
    • one year ago
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    wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice so it would be sec*csc+sec*cot as the numerator

  28. Nnesha
    • one year ago
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    yes right sec*cot + sec *cot = 2 sec*cot \[\huge\rm \frac{ 2\sec*\cot }{ \sec^2 - \cot^2}\] we can convert sec to 1/cos and cot to cos/sin

  29. Nnesha
    • one year ago
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    cot = cos /sin and tan = sin/cos bot hare reciprocal of each other

  30. Nnesha
    • one year ago
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    oh well latex doesn't work |dw:1444619074859:dw|

  31. Nnesha
    • one year ago
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    use the identity cot^2 (x) +1 = csc^2 move the cot^2 to the right side what would you get ?

  32. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }\] simplify the numerator if we solve cot^2 theta +1 =csc^2 for 1 we would get csc^2 -cot^2 =1 so you can replace `csc^2 -cot^2` with 1 simplify the numerator

  33. anonymous
    • one year ago
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    okay

  34. Nnesha
    • one year ago
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    http://math2.org/math/trig/identities.htm copy these down on ur notes identities gt g

  35. anonymous
    • one year ago
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    would i distribute. like 2 x the things in the parentheses

  36. anonymous
    • one year ago
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    @jim_thompson5910 could you help with the last bit?

  37. anonymous
    • one year ago
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    @shaik0124

  38. anonymous
    • one year ago
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    what are we trying to solve. I'm confused.

  39. anonymous
    • one year ago
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    is it the picture you posed?

  40. anonymous
    • one year ago
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    posted*

  41. anonymous
    • one year ago
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    yes the very first picture posted. I have to verify the identity and justify each step

  42. anonymous
    • one year ago
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    oh right

  43. anonymous
    • one year ago
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    i was told that i had it mostly right. we were working from the second line or the third post but i think she started having connection issues

  44. anonymous
    • one year ago
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    let me just do it by hand from the beginning. one sec :)

  45. anonymous
    • one year ago
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    okay thank you so much! this is the last problem that I have to fix and weve been working on it for probably an hour

  46. anonymous
    • one year ago
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    hahaa! i'm sure we will get it done!

  47. anonymous
    • one year ago
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    thank you lol im going on about 18 hours doing school work today!! -_-

  48. anonymous
    • one year ago
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    ok sweet, i got it! lets do this

  49. anonymous
    • one year ago
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    alright:)

  50. anonymous
    • one year ago
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    when i approach these question, what i tend to do is solve the massive weird stuff and compare it to the side which is simplified. so in other words, what i would do with this question is to simplify the left hand side and compare it to the right hand side, right?

  51. anonymous
    • one year ago
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    yes makes sense

  52. anonymous
    • one year ago
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    good! so lefts ONLY work with the LHS (left hand side)

  53. anonymous
    • one year ago
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    \[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\]

  54. anonymous
    • one year ago
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    which would cancel each other out right

  55. anonymous
    • one year ago
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    now, my approach would be to get the denominators the same.

  56. anonymous
    • one year ago
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    see how our denominators are a bit different

  57. anonymous
    • one year ago
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    oh yea ones (-) and the others (+)

  58. anonymous
    • one year ago
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    we can only subtract fractions if there denominators are the same

  59. anonymous
    • one year ago
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    so we are going to multiply the first fraction by the second denominator and vide versa...watch \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]

  60. anonymous
    • one year ago
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    sorry i took too long with syntax

  61. anonymous
    • one year ago
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    thants fine im willing to wait to learn

  62. anonymous
    • one year ago
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    pretty much what this is: \[\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }\]

  63. anonymous
    • one year ago
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    right, i just want to make sure you are on the right track

  64. anonymous
    • one year ago
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    got it?

  65. anonymous
    • one year ago
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    yes im doing it right now

  66. anonymous
    • one year ago
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    now remember the technique of difference of two squares rule: \[(X+Y)(X-Y)=X^2-Y^2\]

  67. anonymous
    • one year ago
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    so this means, \[\left[ \csc(\theta)-\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)-\cot ^{2}(\theta)\]

  68. anonymous
    • one year ago
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    right?

  69. anonymous
    • one year ago
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    right I was just typing it in now lol it takes me a little longer to write it out on paper

  70. anonymous
    • one year ago
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    and thats our denominator right

  71. anonymous
    • one year ago
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    yep! thats our denominator, good!

  72. anonymous
    • one year ago
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    so lets simplify the big equation that i had posted

  73. anonymous
    • one year ago
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    which one?

  74. anonymous
    • one year ago
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    \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\]

  75. anonymous
    • one year ago
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    that one

  76. anonymous
    • one year ago
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    okay that looks better lol

  77. anonymous
    • one year ago
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    omg my syntax keeps screwing up

  78. anonymous
    • one year ago
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    just had to refresh my page haha a

  79. anonymous
    • one year ago
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    \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]

  80. anonymous
    • one year ago
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    good?

  81. anonymous
    • one year ago
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    so lets group everything now

  82. anonymous
    • one year ago
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    \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] -\sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta)}\]

  83. anonymous
    • one year ago
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    expand the numerator

  84. anonymous
    • one year ago
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    \[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\]

  85. anonymous
    • one year ago
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    now simplify! the numerator

  86. anonymous
    • one year ago
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    okay

  87. anonymous
    • one year ago
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    \[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\] this is because sec(theta)csc(theta)-sec(theta)csc(theta)=0

  88. anonymous
    • one year ago
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    so we are left with this

  89. anonymous
    • one year ago
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    so t hey each cancel out leaving us with your fraction above?

  90. anonymous
    • one year ago
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    they*

  91. anonymous
    • one year ago
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    yep!

  92. anonymous
    • one year ago
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    and thats all that I need to do for the question? It doesnt need to be broken down anymore?

  93. anonymous
    • one year ago
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    nope we keep going

  94. anonymous
    • one year ago
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    okay i thought so

  95. anonymous
    • one year ago
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    did nnesha explain to you why \[\csc ^{2}(\theta) -\cot ^{2}(\theta)=1\]

  96. anonymous
    • one year ago
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    if so i dont quite understand still

  97. anonymous
    • one year ago
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    do you have a list of trig identities?

  98. anonymous
    • one year ago
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    yes

  99. anonymous
    • one year ago
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    http://www.purplemath.com/modules/idents.htm where it says \[1 + \cot ^{2}(t) = \csc ^{2}(t)\]

  100. anonymous
    • one year ago
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    so we just re-arrange that equation \[\csc ^{2}(t) - \cot ^{2}(t)=1\]

  101. anonymous
    • one year ago
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    so cot^2(t) is always going to = 1

  102. anonymous
    • one year ago
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    no, \[\csc ^{2}(t) - \cot ^{2}(t)=1\]

  103. anonymous
    • one year ago
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    oh i see what you mean now

  104. anonymous
    • one year ago
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    so where we how we have our denominator as \[\csc ^{2}(t) - \cot ^{2}(t)\] we can just substitute that as 1 since we know the trig identity that \[\csc ^{2}(t) - \cot ^{2}(t)=1\]

  105. anonymous
    • one year ago
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    right?

  106. anonymous
    • one year ago
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    correct

  107. anonymous
    • one year ago
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    awesome :)

  108. anonymous
    • one year ago
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    so that means \[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\] REDUCES to: \[\frac{ 2\sec(\theta)\cot(\theta) }{ 1}\] which is simply \[2\sec(\theta)\cot(\theta) \]

  109. anonymous
    • one year ago
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    understand that?

  110. anonymous
    • one year ago
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    yes because csc^2 (0)-cot^2(0)=1

  111. anonymous
    • one year ago
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    now all we need to know is some basic trig formulas: \[\csc(\theta)=\frac{ 1 }{ \sin (\theta) }\] \[\sec(\theta)=\frac{ 1 }{ \cos(\theta) }\] and \[\cot(\theta)=\frac{ 1 }{ \tan(\theta) }=\frac{ \cos(\theta) }{ \sin(\theta) }\]

  112. anonymous
    • one year ago
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    so \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)\] the cos(X) part cancels since we are multiplying fractions, so we are left with: \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)\]

  113. anonymous
    • one year ago
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    now we know that: \[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\]

  114. anonymous
    • one year ago
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    therefore \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\]

  115. anonymous
    • one year ago
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    now we can see that the LHS reduces to \[2\csc(\theta)\] and that means this is also the RHS!!!!!!

  116. anonymous
    • one year ago
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    which 2 ces(0) is the final answer

  117. anonymous
    • one year ago
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    YEP! so that means we are dun!

  118. anonymous
    • one year ago
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    So for what the teachers asking i simply put everything we have figured out together for her answer

  119. anonymous
    • one year ago
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    so what we did was reduce the left hand side right? and we found that the left hand side equalled to the right hand side

  120. anonymous
    • one year ago
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    yes that was the first step we did

  121. anonymous
    • one year ago
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    Working with the LHS, we want to reduce this to see if it equals the RHS. LHS IMPLIES: \[\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }\] Getting denominators the same: \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}\] Now from denominators, we can use the difference of two squares: \[\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\] Now subtract the two fractions since we have the denominators the same: I.e get it into one fraction. \[\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }\] Simplifying the numerator we get: \[\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}\] Now, using trig identities for the numerator where \[\csc ^{2}(\theta) -\cot ^{2}(\theta)=1\] Therefore, it reduces to \[\frac{ 2\sec(\theta)\cot(\theta) }{ 1}\] Using basic trig identities, we can simplify in terms of sin(x) and cos(x) \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)\] Cancelling out the cos(theta) we get, \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)\] finally we know \[\csc(\theta)=\frac{ 1 }{ \sin(\theta)}\] therefore \[2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)\] =RHS (As required)

  122. anonymous
    • one year ago
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    :)

  123. anonymous
    • one year ago
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    THANK YOU SO MUCH!!! You are a huge lifesaver!!! This was my last question I had to correct and understand for me to complete my diploma. You truly deserve best response for your very detailed help! Thank you thank you!!!

  124. anonymous
    • one year ago
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    : ) you're too kind

  125. anonymous
    • one year ago
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    I gave you best response and a raving testimonial on your page!

  126. anonymous
    • one year ago
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    hahahaha if you want you can fan me so if you stuck with trig equations, you can easily find me if i'm online :) well done though. they are a bit tricky, but there is always a systematic approach!

  127. anonymous
    • one year ago
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    Im already a fan(; If i have any other questions I will be sure to ask you! thank you again and have a wonderful night/morning (its 1am here)

  128. anonymous
    • one year ago
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    jeez. i live in australia so its 3.30pm here haha

  129. anonymous
    • one year ago
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    im in michigan,usa lol

  130. anonymous
    • one year ago
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    i should be doing lecture writing, but i procrastinate on open study

  131. anonymous
    • one year ago
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    ahah jeez. thats hectic

  132. anonymous
    • one year ago
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    see im good at writting so if you need help with THAT let me know lol I will be on tomorrow but right now..im off to bed! my two year old wakes up too early lol

  133. anonymous
    • one year ago
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    haha good night

  134. anonymous
    • one year ago
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    goodnight(;

  135. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }\] simplify the numerator if we solve cot^2 theta +1 =csc^2 for 1 we would get csc^2 -cot^2 =1 so you can replace `csc^2 -cot^2` with 1 simplify the numerator \(\color{blue}{\text{End of Quote}}\) we replace sec with 1/sec and cot with cos/sin se cancel out the cso theta \[\huge\rm \frac{ 2(\frac{1}{\cancel{cos}}*\frac{\cancel{cos}}{\sin}) }{ \sec^2 \theta -\cot^2 }\] and csc =1/sin like i said they are reciprocal as mentioned above sec^2 +1 = cot^2 set it equal to one so we can replace sec^2 -cot^2 with 1

  136. Nnesha
    • one year ago
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    and yea sorry *net issues* :/

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