A) \[-\sec \theta +\tan \theta \]
B) \[\sec \theta +\tan \theta \]
C) \[\sec \theta -\tan \theta \]
D) None of these
Correct Answer: A
Solution :
\[\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sqrt{\frac{{{(1-\sin \theta )}^{2}}}{1-{{\sin }^{2}}\theta }}\] \[=\frac{1-\sin \theta }{\sqrt{{{\cos }^{2}}\theta }}=\frac{1-\sin \theta }{|\cos \theta |}\] \[=\frac{1-\sin \theta }{-\cos \theta }\] \[\left( \because \,\,\frac{\pi }{2}<\theta <\frac{3\pi }{2} \right)\] \[=-\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }\] \[=-\sec \theta +\tan \theta \]
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