<s>[INST] Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$ 
What is 
  $$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$
  for a triangle with sides $2$, $3$, and $4$? 

One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc)$ to get the cosines. But that's lots of calculate.

Is there a better way to get the answer? Thanks!

 [/INST] There are well-known identities for $\triangle ABC$
with the angles $A,B,C$,
sides $a,b,c$, 
semiperimeter $\rho=\tfrac12(a+b+c)$,
area $S$,
radius $r$ of inscribed and 
radius $R$ of circumscribed circles,
\begin{align}
\sin A+\sin B+\sin C
&=\frac\rho{R}
\tag{1}\label{1}
,\\
\cos A+\cos B+\cos C
&=\frac{r+R}{R}
\tag{2}\label{2}
,
\end{align} 
so
\begin{align}
x&=
\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}
=\frac{\rho}{r+R}
\tag{3}\label{3}
,
\end{align} 
we also know that
\begin{align} 
R&=\frac{abc}{4S}
,\\
r&=\frac{S}{\rho}
,\\
S&=\tfrac14\sqrt{4(ab)^2-(a^2+b^2-c^2)^2}
,\\
\end{align} 
thus we can find that for $a=2,b=3,c=4$
\begin{align} 
\rho&=\frac{9}{2}
,\\
S&=\frac{3\sqrt{15}}{4}
,\\
R&=\frac{8\sqrt{15}}{15}
,\\
r&=\frac{\sqrt{15}}{6}
,\\
x&=\frac{\rho}{r+R}
=\frac{3\sqrt{15}}{7}
\approx 1.6598500
.
\end{align}
</s>