# Is $\arccos\left (\frac{\sqrt 6 +1 }{2\sqrt3}\right)= \arctan\left(\frac{\sqrt3 - \sqrt 2}{1+\sqrt 6} \right)$?

Is $\arccos\left(\dfrac{\sqrt 6 +1}{2\sqrt3}\right)= \arctan\left(\dfrac{\sqrt3 - \sqrt 2}{1+\sqrt 6}\right)?$

They are equal according to the calculator but how?

I made a triangle with base $\sqrt 6+ 1$ and hypotenuse $2\sqrt 3$.

Then the height comes out to be $\sqrt{5- 2\sqrt 6}$ using Pythagoras theorem.

But it should come out to be $\sqrt 3 - \sqrt 2$

All you did is fine. Now, notice that$$\left(\sqrt3-\sqrt2\right)^2=3+2-2\sqrt6=5-2\sqrt6.$$