每日一题[1007]消元与求值

若$a\sin^2\theta+b\cos^2\theta=c$且$\dfrac{a}{\sin^2\theta}+\dfrac{b}{\cos^2\theta}=c$，则$\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}$的值为______．

正确答案是$0$．

分析与解　根据已知，有

$$\begin{aligned}a\sin^2\theta+b\cos^2\theta&=c\left(\sin^2\theta+\cos^2\theta\right),\\a\cos^2\theta+b\sin^2\theta&=c\sin^2\theta\cos^2\theta.\end{aligned}$$ 第一个等式即 $$c-b=\dfrac{\sin^2\theta}{\cos^2\theta}\cdot (a-c),$$ 因此 $$\begin{split}&\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\\=&\dfrac{c}{(a-c)+(c-b)}-\dfrac{a}{c-b}-\dfrac{b}{a-c}\\=&\dfrac{c}{\left(1+\dfrac{\sin^2\theta}{\cos^2\theta}\right)\cdot (a-c)}-\dfrac{a\cos^2\theta}{\sin^2\theta\cdot (a-c)}-\dfrac{b\sin^2\theta}{\sin^2\theta\cdot (a-c)}\\=&\dfrac{c\sin^2\theta\cos^2\theta-a\cos^2\theta-b\sin^2\theta}{\sin^2\theta\cdot (a-c)}\\=&0.\end{split}$$

另法　也可以由第一个等式解出$\sin^2\theta=\dfrac {c-b}{a-b}$，代入第二个等式得到

$$\dfrac a{\frac{c-b}{a-b}}+\dfrac b{1-\frac{c-b}{a-b}}=c,$$ 整理得 $$\dfrac{a(a-b)}{c-b}+\dfrac {b(a-b)}{a-c}=c,$$ 两边同除$a-b$得到所求代数式的值为$0$．