Answer:
Option D
Explanation:
We have
$ cosec \theta - \cot \theta=2017$ ........(i)
$\therefore$ $cosec \theta +\cot \theta=\frac{1}{2017}$ ......(ii)
$\begin{bmatrix}\because & cosec^{2}\theta-\cot^{2}\theta=1 \\\Rightarrow & cosec \theta-\cot \theta=\frac{1}{cosec \theta+\cot \theta} \end{bmatrix}$
Adding Eqs,(i) and (ii) , we get
$2 cosec \theta=2017+ \frac{1}{2017}$
$\Rightarrow$ $cosec \theta= \frac{1}{2} [2017+ \frac{1}{2017}]>0$
$\theta$ lie in 1st or II nd quadrant
Substracting Eq.(i) from Eq.(ii) , we get
$2\cot \theta= \frac{1}{2017}-2017$
$\cot \theta= \frac{1}{2}\left(\frac{1}{2017}-2017\right)<0$
$\theta$ lie in II nd and III rd quadrant
Hence , $\theta$ lies in II nd quadrant
