# If $\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.

If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.

I tried$$\sin(\alpha+\beta)=-\sin \beta \cos \beta,\\ 2\sin(\alpha+\beta)=-\sin (2\beta),$$ and $$\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha} =\frac{\sin\alpha\cos^3\beta+\sin^3\beta \cos \alpha}{\sin \alpha \sin \alpha},$$ but unable to find that ratio.

$\def\peq{\mathrel{\phantom{=}}{}}$Denote $γ = α + β$, then$$\frac{\cos α}{\cos β} + \frac{\sin α}{\sin β} = -1 \Longrightarrow \sin γ = \sin(α + β) = -\sin β\cos β.$$ Thus,$$\sin(γ - β) = \sin γ\cos β - \cos γ\sin β = -\sin β\cos^2 β - \cos γ\sin β,\\ \cos(γ - β) = \cos γ\cos β + \sin γ\sin β = \cos γ \cos β - \sin^2 β\cos β.$$ Now, note that$$1 = (\sin^2 β + \cos^2 β)^2 \Longrightarrow \sin^4 β + \cos^4 β = 1 - 2\sin^2 β\cos^2 β,$$ then\begin{align*} &\peq \frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = \frac{\cos^3 β}{\cos(γ - β)} + \frac{\sin^3 β}{\sin(γ - β)}\\ &= \frac{\cos^3 β}{\cos γ \cos β - \sin^2 β\cos β} + \frac{\sin^3 β}{-\sin β\cos^2 β - \cos γ\sin β}\\ &= \frac{\cos^2 β}{\cos γ - \sin^2 β} - \frac{\sin^2 β}{\cos γ + \cos^2 β}\\ &= \frac{\cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)}{(\cos γ - \sin^2 β)(\cos γ + \cos^2 β)}. \tag{1} \end{align*} Because\begin{align*} &\peq \cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)\\ &= (\cos^2 β - \sin^2 β)\cos γ + \sin^4 β + \cos^4 β\\ &= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β, \end{align*} and\begin{align*} &\peq (\cos γ - \sin^2 β)(\cos γ + \cos^2 β)\\ &= \cos^2 γ + (\cos^2 β - \sin^2 β)\cos γ - \sin^2 β\cos^2 β\\ &= (1 - \sin^2 γ) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\ &= (1 - \sin^2 β\cos^2 β) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\ &= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β, \end{align*} then $(1) = 1$, i.e.$$\frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = 1.$$
i only will consider the numerator of your last term: we Can write $$\cos^2(\beta)\sin(\alpha)\cos(\beta)+\sin^2(\beta)\cos(\alpha)\sin(\beta)=$$ $$(1-\sin^2(\beta))\sin(\alpha)\cos(\beta)+(1-\cos^2(\beta))\cos(\alpha)\sin(\beta)=$$ $$\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)-\sin(\alpha)\sin^2(\beta)\cos(\beta)-\cos(\alpha)\cos^2(\beta)\sin(\beta)=$$ $$\sin(\alpha+\beta)-\sin(\beta)\cos(\beta)(\sin(\alpha)\sin(\beta)+\cos(\alpha)\cos(\beta))$$ Can you proceed?