<s>[INST] High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine: 
Let $\alpha$, $\beta$ and $\gamma$  be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).
 [/INST] Consider the first summand, to begin with:
$$
C=\frac{b-2a\cos\gamma}{a\sin\gamma}
$$
The law of cosines tells you that
$$
c^2=a^2+b^2-2ab\cos\gamma
$$
so we can write
$$
b-2a\cos\gamma=\frac{c^2-a^2}{b}
$$
and so we have
$$
C=\frac{c^2-a^2}{ab\sin\gamma}
$$
The law of sines is
$$
\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R
$$
where $R$ is the radius of the circumscribed circle. In particular,
$$
\sin\gamma=\frac{c}{2R}
$$
and so we have
$$
C=\frac{2R(c^2-a^2)}{abc}
$$
Similarly,
$$
A=\frac{c-2b\cos\alpha}{b\sin\alpha}=\frac{a^2-b^2}{abc}\\
B=\frac{a-2c\cos\beta}{c\sin\beta}=\frac{b^2-c^2}{abc}
$$
and
$$
C+A+B=\frac{c^2-a^2+a^2-b^2+b^2-c^2}{abc}=0
$$
</s>