A) \[\frac{40}{65}\]
B) \[\frac{49}{64}\]
C) \[\frac{45}{54}\]
D) \[\frac{49}{74}\]
Correct Answer: B
Solution :
\[\frac{(1+\sin \,\theta )\,(1-\sin \,\theta )}{(1+\cos \,\theta )\,(1-\cos \,\theta )}=\frac{{{(1)}^{2}}-{{(\sin \,\theta )}^{2}}}{{{(1)}^{2}}-{{(\cos \,\theta )}^{2}}}\] \[=\frac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\] \[=\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }={{\left( \frac{\cos \theta }{\sin \theta } \right)}^{2}}\] \[=\,{{(\cot \,\theta )}^{2}}={{\left( \frac{7}{8} \right)}^{2}}=\frac{49}{64}\]
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