Find the $\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}$ Let $a,b,c$ such
$$a\sin^2{x}+b\cos^2{x}=c,~~~\dfrac{a}{\sin^2{x}}+\dfrac{b}{\cos^2{x}}=c$$
find the value
$$\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}$$

Note that $c-a = b\cos^2x - a(1 - \sin^2x) = (b-a) \cos^2x$.
Similarly, $c - b = (a-b)\sin^2x$, so $b-c = (b-a) \sin^2x$
Hence, $\frac{a}{b-c} = \frac{a}{(b-a) \sin^2x}$, and $\frac{b}{c-a} = \frac{b}{(b-a) \cos^2x}$.
Adding these up, $\frac{a}{b-c} + \frac{b}{c-a} = \frac{1}{b-a} \left(\frac{a}{\sin^2 x} + \frac b{\cos^2 x}\right) = \frac{c}{b-a} = \frac{-c}{a-b}$.
Hence, $\frac{a}{b-c} + \frac{b}{c-a} + \frac c{a-b} = \frac{-c + c}{a-b} = 0$.
