​$$\text{If } \frac{\pi}{2} < \theta < \frac{3\pi}{2}, \text{ then } \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}}$$ 
​

Correct option is A

Given:

​$$\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}}$$ 

Formula used:

​$$\cos \theta = \sqrt{1 - \sin^2 \theta}$$​​

Solution:

​$$\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}}$$ 

multiple and divide by 1- sin$$\theta$$​​

=$$\sqrt{\frac{(1 - \sin \theta)\times(1 - \sin \theta)}{1 + \sin \theta\times(1 - \sin \theta)}}$$  

=$$\frac{1 - \sin \theta}{|\cos \theta|}$$

as given value π/2<θ<3π/2 so cos$$\theta$$ = -cos$$\theta$$ 

​$$\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}\ ) = \tan \theta - \sec \theta$$​​