<s>[INST] Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin. 
 [/INST] Let $\tan{a}=x$, $\tan{b}=y$ and $\tan{c}=z$.
Thus, $$0=\sum_{cyc}\frac{\sin(a-b)}{\sin(a+b)}=\sum_{cyc}\frac{x-y}{x+y}=\frac{\sum\limits_{cyc}(x-y)(x+z)(y+z)}{\prod\limits_{cyc}(x+y)}=$$
$$=\frac{\sum\limits_{cyc}(x-y)(z^2+xy+xz+yz)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x-y)z^2}{\prod\limits_{cyc}(x+y)}=$$
$$=\frac{\sum\limits_{cyc}(x^2y-x^2z)}{\prod\limits_{cyc}(x+y)}=\frac{(x-y)(x-z)(y-z)}{(x+y)(x+z)(y+z)},$$
which gives $$(a-b)(a-c)(b-c)=0$$  and we are done!
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