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aussy123

sec B cot B = sinB/cos²B cosB/sin²B 1/sinB Ive been working this out the way I have all the lessons before this, but I dont get the same answer

  • one year ago
  • one year ago

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  1. tomo
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    sec B cot B sec B = (1/cos(B)) cot B = (1/tan(B)) (1/cos(B))*(1/tan(B)) tan(B) = sin(B)/cos(B) (1/cos(B))*(1/(sin(B)/cos(B))) (1/cos(B))*(cos(B)/sin(B)) 1/sin(B)

    • one year ago
  2. aussy123
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    Ooh, I saw those relations at the top but they didn't use the letters, they used delta and I didn't know whether I was supposed to use it. I will copy this in my notes.

    • one year ago
  3. kausarsalley
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    first of all, you would have to know that, \[\sec \theta = \frac{ 1 }{ \cos \theta } \] \[\cot \theta = \frac{ \cos \theta }{ \sin \theta }\]

    • one year ago
  4. kausarsalley
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    you would have to substitute that in your question and solve it.

    • one year ago
  5. aussy123
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    Thank you both, I think I have it now, if I run into any problems Ill def. message you both.

    • one year ago
  6. kausarsalley
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    |dw:1363388213384:dw|

    • one year ago
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