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Let θ be an angle in quadrant II such that sin

Mathematics
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can you continue from here?
Thanks. I'm not sure what to do from here.

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Other answers:

tan x = opposite / adjacent = -1 / sqrt15 sec x = hypotenuse / adjacent = ?
http://faculty.wlc.edu/buelow/PRC/T-2_1.jpg
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Alright now I see. Thanks to both of you.
yw
Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?
Or better yet, why did the 1 become negative when calculating the tangent?
Because your function is in the second quadrant where sine is positive and cosine is negative.
the whole ratio became negative tangent : opposite = 1 , adjacent = - sqrt15 tan = 1 / -sqrt15 = -1/sqrt15
\[\large \left(-\frac{1}{\sqrt{15}}\right)\]
So, sec = -4/sqrt15?
yes
ahh okay I see now thanks.
x values to the left of y-axis are negative y values below x axis are also negative

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