Question 1031102
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*[tex \Large \theta\ \in\ \left(\frac{3\pi}{2},2\pi\right)] means *[tex \Large \theta\ \in\ \text{QIV}]


In QIV, *[tex \Large \sin\varphi <\ 0], so *[tex \Large \csc] and *[tex \Large \cot] are negative as well.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cot\varphi\ =\ \frac{\cos\varphi}{\sin\varphi}]


So if *[tex \Large \cot\theta\ =\ -8] then *[tex \Large \cos\theta\ =\ -8\sin\theta]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos^2\theta\ =\ 64\sin^2\theta]


Substitute from the Pythagorean Identity:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  1\ -\ \sin^2\theta\ =\ 64\sin^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin^2\theta\ =\ \frac{1}{65}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin\theta\ =\ \pm\frac{1}{\sqrt{65}}]


in general, but recall the restriction to QIV, hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ -\frac{1}{\sqrt{65}}] 


Then, since  cosecant is the reciprocal of sine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\theta\ =\ -\sqrt{65}] 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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