If α and β are acute angles satisfying cos 2 α =

(a) $\sqrt{2} \tan \beta$

Given:

$\cos 2 \alpha=\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}$

$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{(3 \cos 2 \beta-1)-(3-\cos 2 \beta)}{(3 \cos 2 \beta-1)+(3-\cos 2 \beta)} \quad$ (Using componendo and dividendo)

$\Rightarrow \frac{\cos 2 \alpha-1}{\cos 2 \alpha+1}=\frac{4 \cos 2 \beta-4}{2 \cos 2 \beta+2}$

$\Rightarrow-\frac{1-\cos 2 \alpha}{1+\cos 2 \alpha}=\frac{-4(1-\cos 2 \beta)}{2(1+\cos 2 \beta)}$

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

$\Rightarrow \frac{2 \sin ^{2} \alpha}{2 \cos ^{2} \alpha}=\frac{2\left(2 \sin ^{2} \beta\right)}{\left(2 \cos ^{2} \beta\right)}$

$\Rightarrow \tan ^{2} \alpha=2 \tan ^{2} \beta$

$\therefore \tan \alpha=\sqrt{2} \tan \beta$