Determining $a+b+c$ given tangent and some roots

Question

$$\tan63^\circ=\sqrt{\sqrt{a}-\sqrt{b}}+\sqrt{\sqrt{c}-\sqrt{b}}$$

where $a,b,c \in \mathbb Z^+$

Determine $a+b+c$

How would I go about this?

Answer 1

HINT: use that $$\tan(63^{\circ})=1/4\, \left( \sqrt {5}-1 \right) \left( 4+\sqrt {2}\sqrt {5-\sqrt {5} } \right) $$

Answer 2

$$\tan63^{\circ}=\frac{1-\cos126^{\circ}}{\sin126^{\circ}}=\frac{1+\sin36^{\circ}}{\cos36^{\circ}}=\frac{1+2\sin18^{\circ}\cos18^{\circ}}{\cos36^{\circ}}=$$ $$=\frac{1+\frac{\sqrt5-1}{2}\cdot\sqrt{1-\left(\frac{\sqrt5-1}{4}\right)^2}}{\frac{\sqrt5+1}{4}}=\frac{4+\frac{\sqrt5-1}{2}\cdot\sqrt{10+2\sqrt5}}{\sqrt5+1}=$$ $$=\frac{\sqrt5-1}{4}\left(4+\frac{\sqrt5-1}{2}\cdot\sqrt{10+2\sqrt5}\right)=\sqrt{5}-1+\frac{3-\sqrt5}{4}\cdot\sqrt{10+2\sqrt5}=$$ $$=\sqrt{6-2\sqrt5}+\sqrt{\frac{(3-\sqrt5)^2(5+\sqrt5)}{8}}=\sqrt{6-2\sqrt5}+\sqrt{\frac{(7-3\sqrt5)(5+\sqrt5)}{4}}=$$ $$=\sqrt{6-2\sqrt5}+\sqrt{5-2\sqrt5}=\sqrt{\sqrt{36}-\sqrt{20}}+\sqrt{\sqrt{25}-\sqrt{20}},$$ which gives $a=36$, $b=20$ and $c=25$.