Question

If $\alpha and\beta$ are acute satisfying $cos2\alpha =\frac{3cos2\beta -1}{3-cos2\beta }$, then $tan\alpha$ =

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

Answer 1

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

If $\frac{a}{b}=\frac{c}{d}$

then,

$\frac{a+b}{a-b}=\frac{c+d}{c-d}$

Now,

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

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

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

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

$\frac{{\cos }^2\alpha }{-{\sin }^2\alpha }=\frac{2{\cos }^2\beta }{-4{\sin }^2\beta }$

$\frac{{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{{\cos }^2\beta }{2{\sin }^2\beta }$

$\Rightarrow \frac{{\sin }^2\alpha }{{\cos }^2\alpha }=\frac{2{\sin }^2\beta }{{\cos }^2\beta }$

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

$\tan \alpha =\sqrt{2}\tan \beta$ ( $\tan \alpha$ and $\tan \beta$ will be positive bacause $\alpha$ and $\beta$ are acute angles. )