Simplify $\arccos{\sqrt{ 2\over 3}}−\arccos{\frac{\sqrt6+1}{2\sqrt3}}$ 

Prove $$\arccos{\sqrt{ 2\over 3}}−\arccos{\dfrac{\sqrt6+1}{2\sqrt3}} = \dfrac\pi6$$

How to proceed with this question? I have tried changing them to $\arctan$ and applying $\arctan a- \arctan b$ but ended up getting some numbers which cant be simplied further.

Let $$\alpha = \arccos{\sqrt2\over \sqrt3}\;\;\;\;{\rm and}\;\;\;\;\beta =\arccos{\sqrt6+1\over 2\sqrt3}$$
so $$ \cos \alpha = \sqrt{2\over 3}\;\;\;\;{\rm and}\;\;\;\;\cos \beta = {\sqrt{6}+1\over 2\sqrt{3}}$$
and 
 $$ \sin \alpha = \sqrt{1-{2\over 3}} ={1\over \sqrt{3}}\;\;\;\;{\rm and}\;\;\;\;\sin \beta = \sqrt{1-{7+2\sqrt{6}\over 12}} = \sqrt{{5-2\sqrt{6}\over 12}}$$ 
so $$\cos (\alpha -\beta) = \sqrt{2\over 3} {\sqrt{6}+1\over 2\sqrt{3}}+{1\over \sqrt{3}}\sqrt{{5-2\sqrt{6}\over 12}}=$$
$$ = {\sqrt{12}+\sqrt{2}+\sqrt{5-2\sqrt{6}}\over 6} $$
$$ = {2\sqrt{3}+\sqrt{2}+\sqrt{(\sqrt{3}-\sqrt{2})^2}\over 6} $$
$$ = {\sqrt{3}\over 2} $$
so $\alpha -\beta = \pi/6$
