You can rationalize the denominator by multiplying by 1 in the form of the conjugate of the denominator divided by itself.
[tex]\dfrac{1}{1+\sqrt{3}}=\dfrac{1}{1+\sqrt{3}}\cdot \dfrac{1-\sqrt{3}}{1-\sqrt{3}}\\\\=\dfrac{1-\sqrt{3}}{1^{2}-(\sqrt{3})^{2}}=\dfrac{1-\sqrt{3}}{1-3}\\\\=\dfrac{\sqrt{3}-1}{2}[/tex]
This eqvaluates to approximately 0.3660254037844386.
