Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$ Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$
where $R$ is the circumradius of the triangle.
Here is my work:
We know that $A+B+C=180^\circ$, so $C=180^\circ -(A+B)$. Plugging this in, we get that $\sin C=\sin (A+B)$ and $\cos C = -\cos (A+B)$. When we plug this into the equation we get, 
$$\frac{a\sin A+b\sin B+c\sin (A+B)}{a\cos A+b\cos B-c\cos (A+B)}.$$
If we expand out $c\sin (A+B)$ and $c\cos (A+B)$, we get 
$$\frac{\sin A+b \sin B+c \cos A\cos B - c\sin A\sin B}{a\cos A+b\cos B-c\cos A\cos B+c\sin A\sin B}.$$
Using the Extended Law of Sines, we can use $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, and $\sin C=\frac{c}{2R}$.
How can I continue on?

Since the sine theorem implies:
$$\sum_\text{cyc}a\sin A = \frac{1}{2R}\sum_\text{cyc}a^2 \tag{1}$$
we just need to prove:
$$ \sum_\text{cyc} a \cos A = \frac{abc}{2R^2}=\frac{2\Delta}{R}\tag 2$$
that is trivial since twice the (signed) area of the triangle made by $B,C$ and the circumcenter $O$ is exactly $aR\cos A$:

