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?


Hint: Go on

$$\frac { 1-\frac { \sqrt { 3 }  }{ 3 } -\frac { \sqrt { 3 }  }{ 3 } +\frac { 1 }{ 3 }  }{ 1-\frac { 1 }{ 3 }  }=\frac { 1-\frac { \sqrt { 3 }  }{ 3 } -\frac { \sqrt { 3 }  }{ 3 } +\frac { 1 }{ 3 }  }{ 1-\frac { 1 }{ 3 }  }\cdot\frac33=
\frac {3-2\sqrt3+1 }{3-1}=
\frac {4-2\sqrt3}{2}=2-\sqrt3$$

$$\large\tan15^{\circ}=2-\sqrt3$$

