\(=\frac{1}{2+\sqrt{4+2\sqrt{3}}}+\frac{1}{2-\sqrt{4-2\sqrt{3}}}=\frac{1}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{1}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{1}{3+\sqrt{3}}+\frac{1}{3-\sqrt{3}}=\frac{3-\sqrt{3}+3+\sqrt{3}}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{6}{6}=1\)