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Q. If $\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}+\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}=1$, then the value of $\left[\frac{\cos ^{4} \beta}{\cos ^{2} \alpha}+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}\right]$ is (where $[\,\,.\,\,]$ denotes greatest integer function)

Trigonometric Functions

Solution:

Given: $\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}+\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}=1$
Let: $\frac{\cos ^{2} \alpha}{\cos \beta}=\cos \theta\,\,\, \frac{\sin ^{2} \alpha}{\sin \beta}=\sin \theta$
$\cos ^{2} \alpha=\cos \beta \cos \theta\,\,\, \sin ^{2} \alpha=\sin \beta \sin \theta$
$1=\cos (\beta-\theta) \Rightarrow \beta=\theta+2 n \pi$
$\therefore \cos ^{2} \alpha=\cos ^{2} \beta$ and $\sin ^{2} \alpha=\sin ^{2} \beta$
$\therefore \frac{\cos ^{4} \beta}{\cos ^{2} \alpha}+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}=\cos ^{2} \beta+\sin ^{2} \beta=1$