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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sec 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ooh, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first of all, you would have to know that, \[\sec \theta = \frac{ 1 }{ \cos \theta } \] \[\cot \theta = \frac{ \cos \theta }{ \sin \theta }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you would have to substitute that in your question and solve it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you both, I think I have it now, if I run into any problems Ill def. message you both.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363388213384:dw
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