<s>[INST] Find the Value of Trigonometric Expression Given $$\begin{align} \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1 \end{align} \tag{1}$$ Find the value of
$$\begin{align} \frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin ^3\beta}{\sin \alpha} \end{align} \tag{2} $$
I Tried like this: From $(1)$ we have$$\sin\alpha \cos\beta+\cos\alpha \sin\beta=-\sin\beta \cos\beta$$ $\implies$
$$\sin(\alpha+\beta)=-\sin\beta \cos\beta \tag{3}$$
Eq $(2)$ is $$\frac{\frac{\sin\alpha}{4}\left(3\cos\beta+\cos3\beta\right)+\frac{\cos\alpha}{4}\left(3\sin\beta-\sin3\beta\right)}{\sin\alpha \cos\alpha}=\frac{\frac{3}{4}\sin(\alpha+\beta)+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}=\frac{\frac{-3}{4}\sin\beta \cos\beta+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}$$
I cannot proceed any further..please help me 
 [/INST] Let $x = \dfrac{\cos \alpha}{\cos \beta}$, and $y = \dfrac{\sin \alpha}{\sin \beta}$, then we have: $x + y = - 1$,and we need to find: $S = \dfrac{\cos ^2\beta}{x} + \dfrac{\sin^2\beta}{y}$.
But $(x\cdot \cos \beta)^2 + (y\cdot \sin \beta)^2 = \cos ^2\alpha + \sin ^2\alpha = 1 \to x^2(1-\sin^2\beta) + y^2(1-\cos^2 \beta) = 1 \to x^2\sin^2\beta + y^2\cos^2\beta = x^2 + y^2 - 1$. Thus:
$S = 1\cdot S = -(x+y)\cdot S = -\left(1 +\dfrac{x}{y}\cdot \sin^2\beta + \dfrac{y}{x}\cdot \cos^2\beta\right) = -\left(1+ \dfrac{x^2+y^2-1}{xy}\right) = -\left(1+\dfrac{1-2xy-1}{xy}\right) = 1$.
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