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 and sines are to be used, your guess is right.
At  first consolidate the cosines by summing up with circular symmetry
$$ c^2=a^2+b^2-2ab\cos\gamma + two more $$  gives
$$ a^2 + b^2 + c^2 = 2 ( b\, c \cos\gamma + ... + ....) \tag {1} $$
For sines consolidation it needs multiplying first term numerator  and denominator by D where  $ c/ \sin \gamma = D, $ the  circumcircle diameter.
The first given term becomes
$$ \dfrac{D}{a\,c}(...) $$
All numerators and all denominators can be added separately and  placed as numerator and denminator of an equivalent fraction  like in:
$$ \frac pq = \frac rs = \frac tu  = \frac {p+r+t}{q+s+u} $$
Combine this result with (1) to simplify it to zero.
