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).

Step 1:
Write cosines relations for a triangle:
$$a^2=b^2+c^2-2bc \cos(A)$$
$$b^2=a^2+c^2-2ac \cos(B)$$
$$c^2=a^2+b^2-2ab \cos(C)$$
Step 2:
Write sinus Area relation for the triangle:
$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$
Step 3:
Write cosines relations as you have in your question:

$$a^2-b^2=c(c-2b \cos(A))$$
$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$

$$b^2-c^2=a(a-2c \cos(B))$$
$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$

$$c^2-a^2=b(b-2a \cos(C))$$
$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
Step 4:
Combine with sinus Area relation 

$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{c b \sin(A) }$$

$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{ac \sin(B)}$$

$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{ab \sin(C)}$$
Step 5:
Combine with sinus Area relation 
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{2.Area(ABC) }$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{2.Area(ABC)}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{2.Area(ABC)}$$
Step 6:
Add them
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$
