<s>[INST] Prove $ \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) $ $ \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) $
This is what I have so far:
I know that $A + B  + C = 180^\circ$, so $C = 180^\circ - (A+B)$. Plugging this in, I get that $\sin C = \sin(A+B)$ and $\cos C = -\cos(A+B)$. When I plug this back into the equation, I get:
$\frac{a\sin A + b\sin B + c \sin(A+B)}{a \cos A + b \cos B - c \cos (A+B)}$.
Expanding out $c \sin(A+B)$ and $c \cos(A+B)$ gives me:
$\frac{a\sin A + b\sin B + c \cos A \cos B - c \sin A \sin B}{a \cos A + b \cos B - c \cos A \cos B + c \sin A \sin B}$.
If I use the Extended Law of Sines, then I get: $ a = 2R * \sin A$, $b = 2R * \sin B$, and $c = 2R * \sin C$, and plugging that in gives me:
$\frac{2R*\sin^{2} A + 2R * \sin^{2} B + 2R \sin C \cos A \cos B - 2R \sin C \sin A \sin B}{2R \sin A \cos A + 2R \sin B \cos B - 2R \sin C \cos A \cos B + 2R \sin C \sin A \sin B}$.
I can factor out all of the $2R's$ to get: 
$\frac{\sin^{2}A + \sin^{2} B + \sin C \cos A \cos B - \sin C \sin A \sin B)}{\sin A \cos A + \sin B \cos B - \sin C \cos A \cos B + \sin C \sin A \sin B}$. 
Now I'm stuck. What do I do next, to get the end result of $R (\frac{a^2+b^2+c^2}{abc})$? 
 [/INST] Your use of Extended Law of Sines is correct. 
Get rid of all the trigo ratios in the numerator.
For the denominator, try to prove that $$a\cos A+b\cos B+c\cos C=R(\sin(2A)+\sin(2B)+\sin(2C))=4\sin A \sin B \sin C$$
Then, use the Extended Law of Sines again! :D
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