A) \[\frac{8}{15}\]
B) \[\frac{15}{8}\]
C) \[\frac{64}{225}\]
D) \[\frac{225}{64}\]
Correct Answer: D
Solution :
\[\frac{(2+2\sin \theta )(1-\sin \theta )}{(1+\cos \theta )\,(2-2cos\theta )}=\frac{2(1+\sin \theta )(1-sin\theta )}{2+(1+\cos \theta )\,(1-cos\,\theta )}\]\[=\frac{(1+\sin \theta )\,(1-\sin \theta )}{(1+\cos \theta )\,(1-\cos \theta )}\] \[=\frac{(1-{{\sin }^{2}}\theta )}{(1-{{\cos }^{2}}\theta )}=\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }={{\cot }^{2}}\theta ={{\left( \frac{15}{8} \right)}^{2}}=\frac{225}{64}\]
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