anemone said:
If $\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$, evaluate $\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}$.
If we let $\sin \alpha=m$, $\sin \beta=n$, $\dfrac{\cos \alpha}{\cos \beta}=p$, the original given equality can rewritten as:
$\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$ $\rightarrow p+\dfrac{m}{n}=-1$
[TABLE="class: grid, width: 800"]
[TR]
[TD]$\therefore \dfrac{m}{n}=-1-p$
$\therefore-n=\dfrac{m}{1+p}$
$\therefore m=-n(1+p)$[/TD]
[TD]$\therefore p+\dfrac{m}{\sin \beta}=-1$
$\sin \beta=-\left( \dfrac{m}{1+p} \right)$
$\sin^2 \beta=\left( \dfrac{m}{1+p} \right)^2$
$\sin^2 \beta=n^2$
$1-\sin^2 \beta=1-n^2$
$\cos^2 \beta=1-n^2$[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]$\therefore \dfrac{\cos^2 \alpha}{\cos^2 \beta}=p^2$
$\dfrac{1-\sin^2 \alpha}{1-\sin^2 \beta}=p^2$
$\dfrac{1-m^2}{1-n^2}=p^2$
$1-m^2=p^2(1-n^2)$
$1-(n(1+p))^2=p^2(1-n^2)$
$p^2+n^2+2pn^2=1$[/TD]
[/TR]
[/TABLE]
We're asked to evaluate
$\begin{align*}
\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}&=\dfrac{\cos^2 \beta }{\dfrac{\cos \alpha}{\cos \beta}}+\dfrac{\sin^2 \beta}{\dfrac{\sin \alpha}{\sin \beta}}\\&=\dfrac{\cos^2 \beta }{p}+\dfrac{\sin^2 \beta}{\dfrac{m}{n}}\\&=\dfrac{\cos^2 \beta }{p}+\dfrac{\sin^2 \beta}{-1-p}\\&=\dfrac{1-n^2 }{p}-\dfrac{n^2}{1+p}\\&=\dfrac{(1-n^2)(1+p)-pn^2}{p(1+p)}\\&=\dfrac{1-n^2+p-pn^2-n^2p}{p(1+p)}\\&=\dfrac{1-(n^2+p^2+2pn^2-p^2)+p}{p(1+p)}\\&=\dfrac{1-1+p^2+p}{p(1+p)}\\&=\dfrac{\cancel{p(1+p)}}{ \cancel{p(1+p)}}\\&=1 \end{align*}$