dontothabonbon
Let θ be an angle in quadrant II such that sin



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cwrw238
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dw:1372535299105:dw

cwrw238
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can you continue from here?

dontothabonbon
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Thanks. I'm not sure what to do from here.

cwrw238
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tan x = opposite / adjacent = 1 / sqrt15
sec x = hypotenuse / adjacent = ?


jdoe0001
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dw:1372535888225:dw

dontothabonbon
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Alright now I see. Thanks to both of you.

cwrw238
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yw

dontothabonbon
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Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?

dontothabonbon
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Or better yet, why did the 1 become negative when calculating the tangent?

Jhannybean
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Because your function is in the second quadrant where sine is positive and cosine is negative.

cwrw238
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the whole ratio became negative
tangent :
opposite = 1 , adjacent =  sqrt15
tan = 1 / sqrt15 = 1/sqrt15

Jhannybean
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\[\large \left(\frac{1}{\sqrt{15}}\right)\]

dontothabonbon
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So, sec = 4/sqrt15?

cwrw238
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yes

dontothabonbon
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ahh okay I see now thanks.

cwrw238
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x values to the left of yaxis are negative
y values below x axis are also negative