Question

If $\alpha ,\beta$ are positive acute angles and ${\cos }^2\alpha =\frac{3\cos 2\beta -1}{3-\cos 2\beta }$, then $\tan \alpha =k\tan \beta$ such that

• A. $k=-\sqrt{2}$
• B. $k=\sqrt{2}$
• C. $k=1$
• D. $k=\sqrt{3}$

Answer 1

Answer: (B)

$k=\sqrt{2}$

$\cos \left(2\alpha \right)=\frac{3\cos 2\beta -1}{3-\cos 2\beta }$
Or
$\frac{1-\cos 2\alpha }{1+\cos 2\alpha }=\frac{(3-\cos 2\beta )-(3\cos 2\beta -1)}{(3-\cos 2\beta )+(3\cos 2\beta -1)}$
Or
${\tan }^2\alpha =\frac{4(1-\cos 2\alpha )}{2(1+\cos 2\alpha )}$
Or
${\tan }^2\alpha =2{\tan }^2\beta$
Or
$\tan \alpha =\sqrt{2}tan\beta$. since $0<\alpha ,\beta <{90}^0$.