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).

First lets consider
$$\frac{b-2a\cos\gamma}{a\sin\gamma}$$
The numerator looks similar to the RHS of the cosine law, but not quite. It would be nice to see $-2ab\cos\gamma$ instead of $-2a\cos\gamma$. So lets just multiply by $b$. Then the numerator would look like this
$$b^2-2ab\cos\gamma$$
Now all that is missing is the $a^2$, so just add it
$$a^2+b^2-2ab\cos\gamma$$
Now my expression is equal to $c^2$. Easy, right?
It would be nice if math was really like this, but unfortunately, we can't just add and multiply arbitrary constants whenever it seems convenient to do so. We can, however, add zero (e.g. $+\phi-\phi$) and multiply by one (e.g. $\frac{\phi}{\phi}$).
$$\frac{b-2a\cos\gamma}{a\sin\gamma} = \frac{b-2a\cos\gamma}{a\sin\gamma}\cdot\frac{b}{b}=\frac{b^2-2ab\cos\gamma}{ab\sin\gamma}=\frac{a^2+b^2-2ab\cos\gamma-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}$$
We have just added zero ($a^2-a^2$), so next, lets multiply by one ($\frac{c}{c}$), so we can isolate $\frac{c}{\sin\gamma}$.
$$\frac{c^2-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}\cdot\frac{c}{c}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}$$
Keeping in mind, of course that
$$\Psi=\frac{c}{\sin\gamma}=\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$$
Then our expression reduces to
$$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}+\frac{a^2-b^2}{abc}\cdot\frac{a}{\sin\alpha}+\frac{b^2-c^2}{abc}\cdot\frac{b}{\sin\beta}$$
$$=\frac{\Psi}{abc}\big((c^2-a^2)+(a^2-b^2)+(b^2-c^2)\big)=0$$
