If operatorname{cosec} theta - cot theta = 20 | vedclass

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(D) We are given that $\operatorname{cosec} 	heta - \cot 	heta = 2017$ $(i)$.Using the identity $\operatorname{cosec}^2 	heta - \cot^2 	heta = 1$,

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$II$

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(D) We are given that $\operatorname{cosec} 	heta - \cot 	heta = 2017$ $(i)$.Using the identity $\operatorname{cosec}^2 	heta - \cot^2 	heta = 1$,we can write $(\operatorname{cosec} 	heta - \cot 	heta)(\operatorname{cosec} 	heta + \cot 	heta) = 1$.Therefore,$\operatorname{cosec} 	heta + \cot 	heta = \frac{1}{2017}$ (ii).Adding $(i)$ and (ii):$2 \operatorname{cosec} 	heta = 2017 + \frac{1}{2017} > 0$,which implies $\operatorname{cosec} 	heta > 0$.Since $\operatorname{cosec} 	heta > 0$,$	heta$ must lie in the $I$ or $II$ quadrant.Subtracting $(i)$ from (ii):$2 \cot 	heta = \frac{1}{2017} - 2017 Since $\cot 	heta Since $	heta$ must satisfy both conditions ($\operatorname{cosec} 	heta > 0$ and $\cot 	heta < 0$),$	heta$ lies in the $II$ quadrant.

If $\operatorname{cosec} \theta - \cot \theta = 2017$,then the quadrant in which $\theta$ lies is

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