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 

Using Holder,s Inequality
$$\displaystyle \bigg[\frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin^3 \beta}{\sin \alpha}\bigg]^2 \cdot (\cos^2 \alpha + \sin^2 \alpha) \geq \bigg[\bigg(\frac{\cos^3 \beta}{\cos \alpha} \cdot \frac{\cos^3 \beta}{\cos \alpha} \cdot \cos^2 \alpha\bigg)^{\frac{1}{3}}+\bigg(\frac{\sin ^3 \beta}{\sin \alpha} \cdot \frac{\sin^3 \beta}{\sin \alpha} \cdot \sin^2 \alpha\bigg)^{\frac{1}{3}}\bigg]^{3} = (\cos^2 \beta+\sin^2 \beta)^3$$
So $$\frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin^3 \beta}{\sin \alpha}\geq 1$$
and equality hold when $$\frac{\cos^3 \beta}{\cos^2 \alpha} = \frac{\sin^3 \beta}{\sin^2 \alpha}\Leftrightarrow \frac{\cos^3  \alpha}{\cos^2 \beta} = \frac{\sin^3 \alpha}{\sin^3 \beta}$$
So $$\frac{\sin^2 \alpha}{\sin^2 \beta}+\frac{\cos^2 \alpha}{\cos^2 \beta}+\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} =0$$
