Question

If $\cos 2\alpha =\frac{3\cos 2\beta -1}{3-\cos 2\beta }$, then $\tan \alpha =\sqrt{2}\tan \beta$

Answer 1

We know that $2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
Using this formula, we get
$\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{3.\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }-1}{3-\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }}=\frac{1-2{\tan }^2\beta }{1+2{\tan }^2\beta }$
Applying componendo and dividendo, we get
$\frac{2{\tan }^2\alpha }{2}=\frac{4{\tan }^2\beta }{2}\therefore {\tan }^2\alpha =2{\tan }^2\beta$
or $\tan \alpha =\sqrt{2}\tan \beta$