Trigonometric identity involving double angles If $\alpha$ and $\beta$ are acute angles and $\displaystyle{\cos2\alpha=\frac{3\cos\beta-1}{3-\cos2\beta}}$, then prove that $\displaystyle{\tan\alpha=\sqrt{2}\tan\beta}$.
I tried this question by taking the formula of $\cos2\alpha$ in terms of $\tan$ (which is of degree two) but I couldn't prove it. Please suggest some hints.
 A: The problem seems to be false. If those equations are true, then these two equations should be equivalent to each other:
$$\alpha  = \frac{1}{2}\arccos \left( {\frac{{3\cos \beta  - 1}}{{3 - \cos 2\beta }}} \right)$$
$$\alpha  = \arctan \left( {\sqrt 2 \tan \beta } \right)$$
But when I graph those equations for varying $\beta$ the graphs are not at all equal.

A: HINT
If I assume              cos B is  cos 2B
then write cos2B= 1 - 2 {sin(B) }^2
then divide both numerator and denominator by {cos^2 (B)}^2 and then use (secB)^2 = 1 + (tanB)^2 
A: Using Weierstrass substitution in either sides.
$$\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{3\cdot\dfrac{1-\tan^2\beta}{1+\tan^2\beta}-1}{3-\dfrac{1-\tan^2\beta}{1+\tan^2\beta}}$$
$$\implies\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{2-4\tan^2\beta}{2+4\tan^2\beta}$$
Using Componendo and dividendo, $$\frac{\tan^2\alpha}1=\frac{4\tan^2\beta}2$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\cos\pars{2\alpha}={3\cos\pars{2\beta} - 1 \over 3 - \cos\pars{2\beta}}\quad
     \imp\quad \tan\pars{\alpha} = \root{2}\tan\pars{\beta}:\ {\large ?}}$

$$
\cos\pars{2\alpha}={3\cos\pars{2\beta} - 1 \over 3 - \cos\pars{2\beta}}\
\imp\ {\cos\pars{2\alpha} + 1 \over \cos\pars{2\alpha} - 1}
={\bracks{3\cos\pars{2\beta} - 1} + \bracks{3 - \cos\pars{2\beta}} \over \bracks{3\cos\pars{2\beta} - 1} - \bracks{3 - \cos\pars{2\beta}}}
$$

$$
{2 - 2\sin^{2}\pars{\alpha} \over 2\cos^{2}\pars{\alpha} - 2}
={2\cos\pars{2\beta} + 2 \over 4\cos\pars{2\beta} - 4}
={\cos\pars{2\beta} + 1 \over 2\cos\pars{2\beta} - 2}
={2 - 2\sin^{2}\pars{\beta} \over 4\cos^{2}\pars{\beta} - 4}
$$

$$
{\cos^{2}\pars{\alpha} \over -\sin^{2}\pars{\alpha}}
={\sec^{2}\pars{\beta} - \tan^{2}\pars{\beta} \over 2 - 2\sec^{2}\pars{\beta}}
={1 \over -2\tan^{2}\pars{\beta}}\ \imp\
\boxed{\quad\tan^{2}\pars{\alpha} = 2\tan^{2}\pars{\beta}\quad}
$$

$$
\color{#66f}{\large\tan\pars{\alpha} = \pm\root{2}\tan\pars{\beta}}
$$
A: If the equation is $$\cos 2\alpha = \frac {3\cos 2\beta -1}{3-\cos 2 \beta}$$ Your method can make progress as follows: first let $t=\tan \alpha$ so that $\cos 2\alpha = \cfrac {1-t^2}{1+t^2}$ and clear fractions - (let $c=\cos 2\beta$) so $$(1-t^2)(3-c)=(1+t^2)(3c-1)$$ so that $$4(1-c)=2(1+c)t^2$$ Now use  $c=\cos 2\beta = \cos^2 \beta- \sin^2\beta$ and $1=\cos^2 \beta+ \sin^2\beta$
You will need the condition that the angles are acute to get the right sign for the square root.
