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).
 A: The following general result holds:
$\tan(x_1+x_2+\cdots+x_n)=\dfrac{s_1-s_3+s_5-s_7+\cdots}{1-s_2+s_4-s_6+\cdots}$
where 
$s_1=\tan x_1+\tan x_2 + \cdots+\tan x_n\quad\quad\;\;:\text{$\quad$sum of the tangents taken one at a time}$
 $s_2=\tan x_1\tan x_2+\tan x_1\tan x_3+ \cdots\quad:\text{$\quad$sum of the tangents taken two at a time}$
and so on.
This can be proved by induction.
Thus,
\begin{align}
&\tan 60^\circ=\tan(\underbrace{1^\circ+1^\circ+\cdots+1^\circ}_\text{60 times})\\\\\\
\\=&\dfrac{\displaystyle{60 \choose 1}\tan 1^\circ-{60 \choose 3}\tan^3 1^\circ+\cdots-{60 \choose 59}\tan^{59}1^\circ}
{\displaystyle 1-{60 \choose 2}\tan^2 1^\circ+{60 \choose 4}\tan^4 1^\circ+\cdots+{60 \choose 60}\tan^{60} 1^\circ}\\\\\\\\
\\=&\dfrac{\displaystyle\sum^{29}_{i=0}{60\choose 2i+1}(-1)^i\tan^{2i+1} 1^\circ}
{\displaystyle\sum^{30}_{i=0}{60\choose 2i}(-1)^i\tan^{2i} 1^\circ}.
\end{align}
A: 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 $a=b=32^\circ$
