A) \[x=y\]
B) \[x<y\]
C) \[x>y\]
D) None of these
Correct Answer: A
Solution :
Since \[{{\cos }^{2}}\theta \le 1\] \[{{\sec }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}}\ge 1\Rightarrow 4xy\ge {{(x+y)}^{2}}\Rightarrow {{(x-y)}^{2}}\le 0\] It is possible only when\[x=y\], \[(\because x,\,y\in R)\].
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