Question

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

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

Answer 1

Answer: (B)

$\sqrt{2}$

$\cos 2\alpha =\frac{3\cos 2\beta -1}{3-\cos 2\beta }$
Applying componendo and dividendo
$\frac{\cos 2\alpha +1}{\cos 2\alpha -1}=\frac{3\cos 2\beta -1+3-\cos 2\beta }{3\cos 2\beta -1-3+\cos 2\beta }=\frac{2\cos 2\beta +2}{4\cos 2\beta -4}\Rightarrow \frac{2{\cos }^2\alpha -1+1}{1-2{sin}^2\alpha -1}=\frac{4{\cos }^2\beta -2+2}{4-8{sin}^2\beta -4}\Rightarrow {\cot }^2\alpha =\frac{1}{2}{\tan }^2\beta \Rightarrow \tan \alpha .\cot \beta =\sqrt{2}$