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- anonymous

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

- schrodinger

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- anonymous

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)

- anonymous

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.

- anonymous

first of all, you would have to know that,
\[\sec \theta = \frac{ 1 }{ \cos \theta } \]
\[\cot \theta = \frac{ \cos \theta }{ \sin \theta }\]

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- anonymous

you would have to substitute that in your question and solve it.

- anonymous

Thank you both, I think I have it now, if I run into any problems Ill def. message you both.

- anonymous

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