If a = sqrt{2} + 1 and b = sqrt{2} - 1, then | vedclass

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(B) Given: $a = \sqrt{2} + 1$ and $b = \sqrt{2} - 1.$We need to find the value of $\frac{1}{a + 1} + \frac{1}{b + 1}.$Substitute the values of $a$ and

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$1$

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(B) Given: $a = \sqrt{2} + 1$ and $b = \sqrt{2} - 1.$We need to find the value of $\frac{1}{a + 1} + \frac{1}{b + 1}.$Substitute the values of $a$ and $b$:$\frac{1}{(\sqrt{2} + 1) + 1} + \frac{1}{(\sqrt{2} - 1) + 1} = \frac{1}{\sqrt{2} + 2} + \frac{1}{\sqrt{2}}.$To simplify,rationalize the first term:$\frac{1}{\sqrt{2} + 2} 	imes \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2 - \sqrt{2}}{4 - 2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}.$Now add the second term:$(1 - \frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} = 1.$Thus,the value is $1$.

If $a = \sqrt{2} + 1$ and $b = \sqrt{2} - 1,$ then the value of $\frac{1}{a + 1} + \frac{1}{b + 1}$ will be

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