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aussy123
Group Title
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
aussy123 Group Title
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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tomo Group TitleBest ResponseYou've already chosen the best response.1
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

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
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

kausarsalley Group TitleBest ResponseYou've already chosen the best response.0
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

kausarsalley Group TitleBest ResponseYou've already chosen the best response.0
you would have to substitute that in your question and solve it.
 one year ago

aussy123 Group TitleBest ResponseYou've already chosen the best response.0
Thank you both, I think I have it now, if I run into any problems Ill def. message you both.
 one year ago

kausarsalley Group TitleBest ResponseYou've already chosen the best response.0
dw:1363388213384:dw
 one year ago
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