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).
The Law of Cosines is equivalent to (and is often proven via) the statements
$$a = b \cos\gamma + c\cos\beta \qquad b = c \cos\alpha + a \cos\gamma \qquad c = a \cos\beta + b \cos\alpha$$
so, your sum becomes
$$\frac{c \cos\alpha - a \cos\gamma}{a\sin\gamma} + \frac{a \cos\beta - b \cos\alpha}{b\sin\alpha} + \frac{b \cos\gamma - c \cos\beta}{c\sin\beta}$$

Further, the Law of Sines allows us to write
$$a = d \sin\alpha \qquad b = d\sin\beta \qquad c = d \sin\gamma$$
where $d$ is the triangle's circumdiameter. This gives

$$\frac{\sin\gamma \cos\alpha - \sin\alpha \cos\gamma}{\sin\alpha\sin\gamma} + \frac{\sin\alpha \cos\beta - \sin\beta \cos\alpha}{\sin\beta\sin\alpha} + \frac{\sin\beta \cos\gamma - \sin\gamma \cos\beta}{\sin\gamma\sin\beta}$$
$$= \left( \cot\alpha - \cot\gamma \right) + \left( \cot\beta - \cot\alpha \right) + \left( \cot\gamma - \cot\beta \right ) = 0 $$
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

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$

$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$