<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] Using an application of the Inscribed Angle Theorem, we get
$$
\begin{align}
2R\sin(A)=a\tag{1a}\\
2R\sin(B)=b\tag{1b}\\
2R\sin(C)=c\tag{1c}
\end{align}
$$
where $R$ is the radius of the circumcircle.
Furthermore, with $s=\frac{a+b+c}2$,
$$
\begin{align}
\text{Area}
&=\sqrt{s(s-a)(s-b)(s-c)}\tag2\\[3pt]
&=\frac12bc\sin(A)\tag3\\
&=\frac{abc}{4R}\tag4
\end{align}
$$
Explanation:
$(2)$: Heron's Formula
$(3)$: triangular area given by Cross Product
$(4)$: apply $\text{(1a)}$ to $(3)$
Therefore,
$$
\begin{align}
\sin(A)+\sin(B)+\sin(C)
&=\frac{a+b+c}{2R}\tag5\\
&=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{abc}\tag6
\end{align}
$$
Explanation:
$(5)$: apply $\text{(1a)}$, $\text{(1b)}$, and $\text{(1c)}$
$(6)$: get $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$ from $(2)$ and $(4)$
The Law of Cosines says
$$
\begin{align}
\cos(A)&=\frac{b^2a+c^2a-a^3}{2abc}\tag{7a}\\
\cos(C)&=\frac{c^2b+a^2b-b^3}{2abc}\tag{7b}\\
\cos(C)&=\frac{a^2c+b^2c-c^3}{2abc}\tag{7c}
\end{align}
$$
Adding these and factoring yields
$$
\begin{align}
\cos(A)+\cos(B)+\cos(C)
&=\frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc}+1\tag8\\
&=\frac{4(s-a)(s-b)(s-c)}{abc}+1\tag9
\end{align}
$$
Combining $(6)$ and $(9)$ gives
$$
\frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)}=\frac{4s\sqrt{s(s-a)(s-b)(s-c)}}{4(s-a)(s-b)(s-c)+abc}\tag{10}
$$

Plugging $(a,b,c)=(2,3,4)$ into $(10)$ gives
$$
\begin{align}
\frac{\sin(A)+\sin(B)+\sin(C)}{\cos(A)+\cos(B)+\cos(C)}
&=\frac{4\cdot\frac92\sqrt{\frac92\cdot\frac52\cdot\frac32\cdot\frac12}}{4\cdot\frac52\cdot\frac32\cdot\frac12+2\cdot3\cdot4}\\
&=\frac{3\sqrt{15}}{7}\tag{11}
\end{align}
$$
</s>