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?

Use Law of Sines for the numerator, $\displaystyle\sum_\text{cyc}a\sin A=\dfrac{\sum_\text{cyc} a^2}{2R}$
For the denominator, $\displaystyle\sum_\text{cyc}a\cos A=\sum_\text{cyc}(2R\sin A\cos A)=R\sum_\text{cyc}\sin2A$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle ,
$\displaystyle\implies\sum_\text{cyc}a\cos A=R(4\sin A\sin B\sin C)=4R\prod_\text{cyc}\left(\dfrac a{2R}\right)=\cdots$
Can you take it home from here?
