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

$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 $
$\Rightarrow \cos ^2 \alpha+\cos ^2 2 \beta-1=\cos ^2 \alpha \cos 2 \beta$
$\Rightarrow \cos ^2 \alpha(1-\cos 2 \beta)=(1-\cos 2 \beta)(1+\cos 2 \beta)=2 \cos ^2 \beta$
$\Rightarrow \frac{\cos ^2 \alpha}{\cos ^2 \beta}=2$