If a(2+sqrt{3})=b(2-sqrt{3})=1, then the valu | vedclass

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(D) Given $a(2+\sqrt{3})=1 \implies a = \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$.Given $b(2-\sqrt{3})=1 \implies b = \frac{1}{2-\sqrt{3}} = 2+\sqrt{3}$.Now,

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(D) Given $a(2+\sqrt{3})=1 \implies a = \frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$.Given $b(2-\sqrt{3})=1 \implies b = \frac{1}{2-\sqrt{3}} = 2+\sqrt{3}$.Now,$a^2 = (2-\sqrt{3})^2 = 4+3-4\sqrt{3} = 7-4\sqrt{3}$.And $b^2 = (2+\sqrt{3})^2 = 4+3+4\sqrt{3} = 7+4\sqrt{3}$.We need to find $\frac{1}{a^2+1} + \frac{1}{b^2+1}$.Substituting the values: $\frac{1}{7-4\sqrt{3}+1} + \frac{1}{7+4\sqrt{3}+1} = \frac{1}{8-4\sqrt{3}} + \frac{1}{8+4\sqrt{3}}$.Taking the common denominator: $\frac{(8+4\sqrt{3}) + (8-4\sqrt{3})}{(8-4\sqrt{3})(8+4\sqrt{3})}$.$= \frac{16}{64 - (16 	imes 3)} = \frac{16}{64-48} = \frac{16}{16} = 1$.

If $a(2+\sqrt{3})=b(2-\sqrt{3})=1,$ then the value of $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}$ is

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