Discussion Forum

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

Answer:

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