If a^{2}+b^{2}+frac{1}{a^{2}}+frac{1}{b^{2}}= | vedclass

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(C) Given the equation: $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}=4$We can rearrange the terms as follows:$(a^{2}-2+\frac{1}{a^{2}})+(b^{2}-2+\frac

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$2$

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(C) Given the equation: $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}=4$We can rearrange the terms as follows:$(a^{2}-2+\frac{1}{a^{2}})+(b^{2}-2+\frac{1}{b^{2}})=0$This can be written as:$(a-\frac{1}{a})^{2}+(b-\frac{1}{b})^{2}=0$Since the sum of squares of real numbers is zero only if each individual term is zero:$(a-\frac{1}{a})=0 \Rightarrow a^{2}=1$$(b-\frac{1}{b})=0 \Rightarrow b^{2}=1$Therefore,the value of $a^{2}+b^{2} = 1+1 = 2$.

If $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}=4,$ then the value of $a^{2}+b^{2}$ will be

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