A) \[x(1-y)=1\]
B) \[y(1+x)=1\]
C) \[y(1-x)=1\]
D) \[x(1+y)=1\]
Correct Answer: C
Solution :
\[\because \,\,\,\,\,\,\,\,\,\,\,\,\,x=\sum\limits_{n=0}^{\infty }{{{(-{{\tan }^{2}}\theta )}^{n}}}=\frac{1}{1-(-{{\tan }^{2}}\theta )}\] \[=\frac{1}{{{\sec }^{2}}\theta }={{\cos }^{2}}\theta \] and \[y=\sum\limits_{n=0}^{\infty }{{{({{\cos }^{2}}\theta )}^{n}}}=\frac{1}{1-{{\cos }^{2}}\theta }=\frac{1}{{{\sin }^{2}}\theta }\] \[\because \,\,\,\,\,\,\,\,\,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{y}+x=1\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1+xy=y\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y(1-x)=1\]
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