Question 1003999
I'm thinking you want the multiplicative inverse (the additive inverse is just -sqrt(3) + 1)...so
{{{1 / (sqrt(3) - 1)}}} 
We multiply top and bottom by the conjugate sqrt(3) + 1 and get
{{{(sqrt(3)+1) / (3 - 1)}}} =   (remember (a+b)(a-b) = a^2 - b^2)
{{{(sqrt(3)+1) / 2}}}