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.