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anonymous
 3 years ago
Let θ be an angle in quadrant II such that sin
anonymous
 3 years ago
Let θ be an angle in quadrant II such that sin

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cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.2can you continue from here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks. I'm not sure what to do from here.

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.2tan x = opposite / adjacent = 1 / sqrt15 sec x = hypotenuse / adjacent = ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372535888225:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Alright now I see. Thanks to both of you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or better yet, why did the 1 become negative when calculating the tangent?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Because your function is in the second quadrant where sine is positive and cosine is negative.

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.2the whole ratio became negative tangent : opposite = 1 , adjacent =  sqrt15 tan = 1 / sqrt15 = 1/sqrt15

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large \left(\frac{1}{\sqrt{15}}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahh okay I see now thanks.

cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.2x values to the left of yaxis are negative y values below x axis are also negative
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