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
$$\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}$$

a$\cos^2x$+b$\sin^2x =c$
Divide this equation by $\cos^2x$.
We have thus,a+b $\tan^2x$=c $\sec^2x$.
$\sec^2x$=$\tan^2x$ +1.This gives the value of $\tan^2x$=(a-c)/(c-b).
Thus values of $\cos^2x$  and $\sin^2x$ are obtained. these values are put in equation a/$\cos^2x$ +  b/$\sin^2x$=c.
Thus the required value is obtained as c/(a-b) + a/(b-c) + b/(c-a)=0.
Hope u understand.
