Express tan(60 deg) as sum of tan(1 deg) Using the identity $tan(a+b)= \frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$, how do you express $tan(60^{\circ})$ in terms of $tan(1^{\circ})$?
Edit: I see that $tan(2^{\circ}+1^{\circ})=\frac{tan(1^{\circ})+tan(2^{\circ})}{1-tan(1^{\circ})tan(2^{\circ})}$ and then we can replace $tan(2^{\circ})$ with $tan(1^{\circ}+1^{\circ})$. It is that I am having a hard time seeing how it iterates (or where the "..."s go).

HINT:
As $60=2^5+2^4+2^3+2^2=2^6-2^2$
First set $a=b=1$ to find $\tan2^\circ$ in terms of $\tan1^\circ$
Then set $a=b=2^\circ$ to find $\tan4^\circ$ in terms of $\tan2^\circ$ which is already in terms of $\tan1^\circ$
Then $a=b=4^\circ$
Then $a=b=8^\circ$
Then $a=b=16^\circ$

or $60=2^6-2^2$
Then $a=b=32^\circ$
