Evaluting $\tan15°$ using difference formula 
Evalute $\tan15°$ using difference formula

Steps I took:
$$\begin{align}
\tan(45-30)&=\frac { \tan(45)-\tan(30) }{ 1+\tan(45)\tan(30) }\\
&=\frac { 1-\frac { \sqrt { 3 }  }{ 3 }  }{ 1+\frac { \sqrt { 3 }  }{ 3 }  } \\
&=\frac { 1-\frac { \sqrt { 3 }  }{ 3 }  }{ 1+\frac { \sqrt { 3 }  }{ 3 }  } \cdot \frac { 1-\frac { \sqrt { 3 }  }{ 3 }  }{ 1-\frac { \sqrt3 }{ 3 }  }\\
&=\frac { 1-\frac { \sqrt { 3 }  }{ 3 } -\frac { \sqrt { 3 }  }{ 3 } +\frac { 1 }{ 3 }  }{ 1-\frac { 1 }{ 3 }  } \\
\end{align}$$
Where do I go from here to arrive at the correct answer? And what mistakes did I make, if any?

$$\frac{1-\dfrac1{\sqrt3}}{1+\dfrac1{\sqrt3}}=\frac{\dfrac{\sqrt3-1}{\sqrt3}}{\dfrac{\sqrt3+1}{\sqrt3}}$$
$$=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}=?$$
