Question

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)

Answer 1

Answer: 1

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