Find the value of $\tan \frac{13 \pi}{12}$

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We have

$\tan \frac{{13\pi }}{{12}} = \tan \left( {\pi  + \frac{\pi }{{12}}} \right)$

$ = \tan \frac{\pi }{{12}} = \tan \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$

$ = \frac{{\tan \frac{\pi }{4} - \tan \frac{\pi }{6}}}{{1 + \tan \frac{\pi }{4}\tan \frac{\pi }{6}}}$

$ = \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} = \frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}$

$= 2 - \sqrt 3 $

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