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.

Hint: show that $\cos(\text{your expression})$ simplifies to what it should.
Then show that your expression is in the correct interval.
