(cos(theta/2)-sqrt(1+sintheta))/(sin(theta/2)-sqrt(1+sintheta))=? 

প্রথমে, দেওয়া সমীকরণটি লিখি:

\[ \frac{\cos(\theta/2) - \sqrt{1 + \sin \theta}}{\sin(\theta/2) - \sqrt{1 + \sin \theta}} = ? \]

প্রথমে, \( \sin \theta \) এর জন্য সম্পর্ক ব্যবহার করি: \[ \sin \theta = 2 \sin(\theta/2) \cos(\theta/2) \] এবং, \[ 1 + \sin \theta = 1 + 2 \sin(\theta/2) \cos(\theta/2) \] এটি লিখতে পারি: \[ 1 + \sin \theta = (\sin(\theta/2) + \cos(\theta/2))^2 \] কারণ, \[ (\sin(\theta/2) + \cos(\theta/2))^2 = \sin^2(\theta/2) + 2 \sin(\theta/2) \cos(\theta/2) + \cos^2(\theta/2) = 1 + 2 \sin(\theta/2) \cos(\theta/2) \] অর্থাৎ, \[ \sqrt{1 + \sin \theta} = \sin(\theta/2) + \cos(\theta/2) \] এখন, মূল সমীকরণে প্রতিস্থাপন করি: \[ \frac{\cos(\theta/2) - (\sin(\theta/2) + \cos(\theta/2))}{\sin(\theta/2) - (\sin(\theta/2) + \cos(\theta/2))} \] সরলীকরণ করি: \[ \frac{\cos(\theta/2) - \sin(\theta/2) - \cos(\theta/2)}{\sin(\theta/2) - \sin(\theta/2) - \cos(\theta/2)} = \frac{- \sin(\theta/2)}{- \cos(\theta/2)} \] এটি সরলীকরণ করলে: \[ \frac{\sin(\theta/2)}{\cos(\theta/2)} = \tan(\theta/2) \] অতএব, উত্তর হলো:

\(\boxed{\tan(\theta/2)}\)