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