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Mathematics 22 Online
OpenStudy (anonymous):

If \(cosec(\theta) + cot(\theta) = 6\), then: \[a) \quad cosec(\theta) = \frac{35}{12}, \quad \cot(\theta) = \frac{37}{12}\] \[b) \quad cosec(\theta) = \frac{37}{12}, \quad \cot(\theta) = \frac{35}{12}\] \[c) \quad cosec(\theta) = \frac{41}{12}, \quad \cot(\theta) = \frac{31}{12}\] \[d) \quad None \; of \: these.. \]

OpenStudy (anonymous):

would this be productive? \[\frac{ 1 }{ \sin \theta }+\frac{ \cos \theta }{ \sin \theta }=6\] \[1+\cos \theta=6\sin \theta\] probably not.

OpenStudy (anonymous):

but this does |dw:1366096794723:dw|

OpenStudy (anonymous):

How you made that triangle and how you calculated those values?

OpenStudy (anonymous):

pythagorean theorem with the multiple choice answers given

OpenStudy (anonymous):

@campbell_st can you solve this by not using answer choices?

OpenStudy (anonymous):

Yes we can use the answer choices but what if the question says that you are given with cosecx + cotx = 6 find cosecx and cot(x), then how will you find it??

OpenStudy (anonymous):

try to find x? I don't know.

OpenStudy (anonymous):

you can probably use a graphing calculator to find x. graph y=6 sin theta - cos theta -1

OpenStudy (anonymous):

and find where y=0

OpenStudy (anonymous):

You are going deeper, I will not prefer this way, never mind.. Is their any simple way?? @hartnn @dumbcow @Luis_Rivera

OpenStudy (anonymous):

that or I would graph cosec x + cot x -6 and find y=0 for x that's probably simpler

OpenStudy (anonymous):

that's the simple way graph y=cosec x + cot x -6 find x with a graphing calculator calculate values of cosec and cot for that x you will get approximate values, but I don't know how to do anything better than that.

OpenStudy (anonymous):

It is just an aptitude type question, you have only 20 seconds to do it.. Can you do it in 20 seconds?? Ha ha h ha..

OpenStudy (anonymous):

only the multiple choice way

OpenStudy (anonymous):

I got the triangle in 20 sec

OpenStudy (anonymous):

Then yes I will prefer that way..

OpenStudy (anonymous):

Thank you very much @Peter14 ..

OpenStudy (anonymous):

once I knew to try that way you're welcome.

OpenStudy (raden):

1+cosθ=6sinθ square both sides 1+2cosθ+cos^2θ=36sin^2θ 1+2cosθ+cos^2θ=36(1-cos^2θ) 37cos^2θ+2cosθ-35=0 factor out! (37cosθ - 35)(cosθ + 1) = 0 solve for cosθ, first

hartnn (hartnn):

^

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