## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$6-3\sqrt 3$
By multiplying both denominator and nominator by $2-\sqrt{3}$, the given expression simplifies to: $=\dfrac{3}{2+\sqrt{3}} \cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}}$ $=\dfrac{3(2-\sqrt{3})}{2^2-\sqrt9}$ $=\dfrac{6-3\sqrt{3}}{4-3}$ $=\dfrac{6-3\sqrt{3}}{1}$ $=6-3\sqrt 3$