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