Question : The value of $\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1-\sqrt{2}+\sqrt{3}}$ is:
Option 1: $\sqrt{2}$
Option 2: $\sqrt{3}$
Option 3: $1$
Option 4: $4(\sqrt{3}+\sqrt{2})$
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Correct Answer: $1$
Solution :
$\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1-\sqrt{2}+\sqrt{3}}$
$= \frac{1}{(1+\sqrt{3})+\sqrt{2}}+\frac{1}{(1+\sqrt{3})-\sqrt{2}}$
$= \frac{1+\sqrt{3}-\sqrt{2}+1+\sqrt{3}+\sqrt{2}}{[(1+\sqrt{3})+\sqrt{2}][(1+\sqrt{3})-\sqrt{2}]}$
$= \frac{2\sqrt{3}+2}{(1+\sqrt{3})^{2}-(\sqrt{2})^{2}}$
$= \frac{2\sqrt{3}+2}{1+3+2\sqrt{3}-2}$
$= \frac{2\sqrt{3}+2}{2\sqrt{3}+2}$
$=1$
Hence, the correct answer is $1$.
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