Determine exact values using angle sum and difference identity (trig) Ok so we have just learnt the basic angle sum and difference identities:
$$\begin{array}{l}\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B\\\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B\\\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\end{array}
$$
The question i have been given is determine the exact value of $\tan {15^0}$ degrees. 
So i assume you use the tan identity, sub in 
$$\tan \left( {{{45}^0} - {{30}^0}} \right) = \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan 30}}
$$
After simplifying it down as far as possible i end up with:
$$\frac{{\left( {3\left( {3 - \sqrt 3 } \right)} \right)}}{{\left( {3\left( {3 + \sqrt 3 } \right)} \right)}}
$$
However the correct answer is: ${2 - \sqrt 3 }$
I'm not sure if iv'e done something wrong or are completely off track, but could someone please explain (as simply as possible) how to achieve this answer. 
 A: You can go further:
\begin{align}\frac{3(3-\sqrt3)}{3(3+\sqrt3)}&=\frac{3-\sqrt3}{3+\sqrt3}\\\\
&=\frac{9-6\sqrt3+3}{3^2-3}\\\\
&=\frac{12-6\sqrt3}6\\\\
&=2-\sqrt3\end{align}
A: You haven't done anything wrong, you just haven't finished yet - it is good practice to not leave irrational numbers ($\sqrt3$) in the denominator.
$$\frac{3-\sqrt{3}}{3+\sqrt3}=\frac{3-\sqrt3}{3+\sqrt3}\cdot\frac{3-\sqrt3}{3-\sqrt3}=\frac{9+3-6\sqrt3}{9-3}=2-\sqrt3$$
A: Here you go, 
$$\begin{align}\tan \left( {{{45}^0} - {{30}^0}} \right) &= \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan {{30}^0}}}\\\\
&= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \left( 1 \right)\left( {\frac{1}{{\sqrt 3 }}} \right)}}\\\\
&= \frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}\\ \\
&= \frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} \times \frac{{\sqrt 3  - 1}}{{\sqrt 3  - 1}}\\\\
&= \frac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}\\\\
&= \frac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\\\\ 
&= \frac{{4 - 2\sqrt 3 }}{2}\\\\ 
&= 2 - \sqrt 3 \end{align}
$$
