Answer
$-\dfrac {1}{2}\left( 5\sqrt {2}+4\sqrt {3}\right) $
Work Step by Step
$\tan 285=\tan \left( 225+60\right) =\dfrac {\tan 225+\tan 60}{1-\tan 225\tan 60}=\dfrac {\dfrac {\sqrt {2}}{2}+\sqrt {3}}{1-\dfrac {\sqrt {2}}{2}\sqrt {3}}=\dfrac {\sqrt {2}+\sqrt {3}}{1-\sqrt {6}}$
$=\dfrac {\left( \sqrt {2}+\sqrt {3}\right) \left( 2+\sqrt {6}\right) }{\left( 2-\sqrt {6}\right) \left( 2+\sqrt {6}\right) }=\dfrac {2\sqrt {2}+2\sqrt {3}+2\sqrt {3}+3\sqrt {2}}{2^{2}-\left( \sqrt {6}\right) ^{2}}=\dfrac {5\sqrt {2}+4\sqrt {3}}{-2}=-\dfrac {1}{2}\left( 5\sqrt {2}+4\sqrt {3}\right) $