If _{$\sqrt{\frac{1+\sin A}{1-\sin A}}=-\left(\sec A+\tan A\right)$}  then the quadrants in which the angle A lies are

$$\begin{gathered} \sqrt{\frac{1+\sin A}{1-\sin A}}=-(\sec A+\tan A) \\ \Rightarrow \sqrt{\frac{\left(1+\sin A\right)\left(1+\sin A\right)}{\left(1-\sin A\right)\left(1+\sin A\right)}}=-(\sec A+\tan A) \end{gathered}$$

  $\Rightarrow \frac{1+\sin A}{\sqrt{{\cos }^{2}A}}=-\left(\sec A+\tan A\right)$

  $\Rightarrow \cos A$  is  $-ve$ 

  $\Rightarrow A\in II,III quadrants$