5. Given that $\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}+\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}=1$, evaluate $\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}$.
5. Answer. 1
Solution. Since $\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}+\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}=1$, set $\cos \theta=\frac{\cos ^{2} \alpha}{\cos \beta}$ and $\sin \theta=\frac{\sin ^{2} \alpha}{\sin \beta}$. Then
$$
\cos (\theta-\alpha)=\cos \theta \cos \alpha+\sin \theta \sin \alpha=\cos ^{2} \alpha+\sin ^{2} \alpha=1 \text {. }
$$
and so
$\theta-\alpha=2 k \pi$ for some $k \in \mathbb{Z}$.

Thus $\sin \theta=\sin \alpha$ and $\cos \theta=\cos \alpha$. Consequently,
$$
\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}=\cos ^{2} \beta+\sin ^{2} \beta=1 .
$$