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 one year 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
 one year 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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tomo
 one year ago
Best ResponseYou've already chosen the best response.1sec 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)

aussy123
 one year 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.

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

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

aussy123
 one year 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.

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