Question

If $\sec (\alpha -2\beta )$, $\sec \alpha$ and $\sec (\alpha +2\beta )$ are distinct terms in arithmetic progression, then find the value of $\frac{{\cos }^2\alpha }{{\cos }^2\beta }$

Answer 1

Answer: 2

$2\sec \alpha =\sec (\alpha -2\beta )+\sec (\alpha +2\beta )$
$\Rightarrow \frac{2}{\cos \alpha }=\frac{1}{\cos (\alpha -2\beta )}+\frac{1}{\cos (\alpha +2\beta )}$
$\Rightarrow \frac{2}{\cos \alpha }=\frac{2\cos \alpha \cdot \cos 2\beta }{\cos (\alpha -2\beta )\cos (\alpha +2\beta )}$
$\Rightarrow \cos 2\alpha +\cos 4\beta =2{\cos }^2\alpha \cos 2\beta$
$\Rightarrow \left(2{\cos }^2\alpha -1\right)+\left(2{\cos }^{2}2\beta -1\right)=2{\cos }^2\alpha \cos 2\beta$
$\Rightarrow {\cos }^2\alpha +{\cos }^{2}2\beta -1={\cos }^2\alpha \cos 2\beta$
$\Rightarrow {\cos }^2\alpha (1-\cos 2\beta )=(1-\cos 2\beta )(1+\cos 2\beta )=2{\cos }^2\beta$
$\Rightarrow \frac{{\cos }^2\alpha }{{\cos }^2\beta }=2$