Question

If $\alpha$ and $\beta$ are acute angles satisfying $\cos 2\alpha =\frac{3\cos 2\beta -1}{3-\cos 2\beta }$, then $\tan \alpha =$.

• A. $\sqrt{2}\tan \beta$
• B. $1/\sqrt{2}\tan \beta$
• C. $\sqrt{2}\cot \beta$
• D. $1/\sqrt{2}\cot \beta$

Answer 1

The correct option is A $\sqrt{2}\tan \beta$

$\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

$\therefore \cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

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

$\therefore \frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{3\cos 2\beta -1}{3-\cos 2\beta }$

$=\frac{3\left(\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }\right)-1}{1-\left(\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }\right)}$

$\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }=3\frac{3-3{\tan }^2\beta -1-{\tan }^2\beta }{3+3{\tan }^2\beta -1+{\tan }^2\beta }$

$=\frac{2-4{\tan }^2\beta }{2+4{\tan }^2\beta }$

$\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{1-2{\tan }^2\beta }{1+2{\tan }^2\beta }$

Componendo and Dividendo

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

$\Rightarrow \frac{2}{2{\tan }^2\alpha }=\frac{2}{4{\tan }^2\beta }$

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

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