A) \[{{G}^{2}}\]
B) \[\frac{1}{{{G}^{2}}}\]
C) \[\frac{2}{{{G}^{2}}}\]
D) \[3{{G}^{2}}\]
Correct Answer: B
Solution :
As given \[G=\sqrt{xy}\] \[\therefore \]\[\frac{1}{{{G}^{2}}-{{x}^{2}}}+\frac{1}{{{G}^{2}}-{{y}^{2}}}=\frac{1}{xy-{{x}^{2}}}+\frac{1}{xy-{{y}^{2}}}\] \[=\frac{1}{x-y}\left\{ -\frac{1}{x}+\frac{1}{y} \right\}=\frac{1}{xy}=\frac{1}{{{G}^{2}}}\].
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