Question

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

• A. $1$
• B. $-1$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

Answer: (C), (D)

The correct options are
C $\sqrt{2}$
D $-\sqrt{2}$

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

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

Applying componendo and dividendo, we get

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