Find the $\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}$

Question

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

Answer 1

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, $\frac{a}{b-c} + \frac{b}{c-a} + \frac c{a-b} = \frac{-c + c}{a-b} = 0$.

Answer 2

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.