A) \[\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }\]
B) \[\frac{1}{1+\cos \theta }-\frac{1}{1-\sin \theta }\]
C) \[\frac{1}{1+\cos \theta }+\frac{1}{1-\sin \theta }\]
D) \[\frac{1}{1-\cos \theta }-\frac{1}{1+\sin \theta }\]
Correct Answer: A
Solution :
Given, \[{{x}^{2}}\sin \theta -x(sin\theta cos\theta +1)+cos\theta =0\] \[D={{(1+sin\theta cos\theta )}^{2}}-4\sin \theta \cos \theta \]\[={{(1-sin\theta cos\theta )}^{2}}\] \[\therefore \]\[x=\frac{1+\sin \theta \cos \theta \pm (1-sin\theta cos\theta )}{2\sin \theta }\] \[\Rightarrow \]\[\alpha =\cos \theta \]and\[\beta =\operatorname{cosec}\theta \] Now,\[\sum\limits_{n=0}^{\infty }{\left( {{\alpha }^{n}}+{{\left( -\frac{1}{\beta } \right)}^{n}} \right)}=\sum\limits_{n=0}^{\infty }{{{(\cos \theta )}^{n}}}+\sum\limits_{n=0}^{\infty }{{{(-sin\theta )}^{n}}}\] \[=\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }\]
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