If \(cosec(\theta) + cot(\theta) = 6\), then: \[a) \quad cosec(\theta) = \frac{35}{12}, \quad \cot(\theta) = \frac{37}{12}\] \[b) \quad cosec(\theta) = \frac{37}{12}, \quad \cot(\theta) = \frac{35}{12}\] \[c) \quad cosec(\theta) = \frac{41}{12}, \quad \cot(\theta) = \frac{31}{12}\] \[d) \quad None \; of \: these.. \]
would this be productive? \[\frac{ 1 }{ \sin \theta }+\frac{ \cos \theta }{ \sin \theta }=6\] \[1+\cos \theta=6\sin \theta\] probably not.
but this does |dw:1366096794723:dw|
How you made that triangle and how you calculated those values?
pythagorean theorem with the multiple choice answers given
@campbell_st can you solve this by not using answer choices?
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??
try to find x? I don't know.
you can probably use a graphing calculator to find x. graph y=6 sin theta - cos theta -1
and find where y=0
You are going deeper, I will not prefer this way, never mind.. Is their any simple way?? @hartnn @dumbcow @Luis_Rivera
that or I would graph cosec x + cot x -6 and find y=0 for x that's probably simpler
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.
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..
only the multiple choice way
I got the triangle in 20 sec
Then yes I will prefer that way..
Thank you very much @Peter14 ..
once I knew to try that way you're welcome.
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
^
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