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

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

dontothabonbon
 one year ago
Best ResponseYou've already chosen the best response.0Thanks. I'm not sure what to do from here.

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

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1dw:1372535888225:dw

dontothabonbon
 one year ago
Best ResponseYou've already chosen the best response.0Alright now I see. Thanks to both of you.

dontothabonbon
 one year 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?

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

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

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

dontothabonbon
 one year ago
Best ResponseYou've already chosen the best response.0So, sec = 4/sqrt15?

dontothabonbon
 one year ago
Best ResponseYou've already chosen the best response.0ahh okay I see now thanks.

cwrw238
 one year 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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