A) \[x=y\]
B) \[x>y\]
C) \[x<y\]
D) None of these
Correct Answer: A
Solution :
Given equation, \[{{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}}\] Since range of \[\sec \theta \] is \[(-\infty ,-1]\,\cup [1,\infty ).\] \[\therefore \,\,{{\sec }^{2}}\theta \ge 1\] \[\Rightarrow \,\,\frac{4xy}{{{(x+y)}^{2}}}\ge 1\] \[\Rightarrow \,\,{{(x-y)}^{2}}\le 0\] But \[{{(x-y)}^{2}}</0\] for any x, \[y\in R\] \[\therefore \,\,\,{{(x-y)}^{2}}=0\Rightarrow x=y\]
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