Question

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

(a) $\sqrt{2}\tan \beta$

$$\begin{gathered} \mathrm{Given}: \\ \cos 2\mathrm{\alpha }=\frac{3\cos 2\mathrm{\beta }-1}{3-\cos 2\mathrm{\beta }} \\ \Rightarrow \frac{\cos 2\mathrm{\alpha }-1}{\cos 2\mathrm{\alpha }+1}=\frac{\left(3\cos 2\mathrm{\beta }-1\right)-\left(3-\cos 2\mathrm{\beta }\right)}{\left(3\cos 2\mathrm{\beta }-1\right)+\left(3-\cos 2\mathrm{\beta }\right)}\left(\mathrm{Using}\mathrm{componendo}\mathrm{and}\mathrm{d}\mathrm{ividendo}\right) \\ \Rightarrow \frac{\cos 2\mathrm{\alpha }-1}{\cos 2\mathrm{\alpha }+1}=\frac{4\cos 2\mathrm{\beta }-4}{2\cos 2\mathrm{\beta }+2} \\ \Rightarrow -\frac{1-\cos 2\mathrm{\alpha }}{1+\cos 2\mathrm{\alpha }}=\frac{-4\left(1-\cos 2\mathrm{\beta }\right)}{2\left(1+\cos 2\mathrm{\beta }\right)} \\ \Rightarrow \frac{1-\cos 2\mathrm{\alpha }}{1+\cos 2\mathrm{\alpha }}=\frac{2\left(1-\cos 2\mathrm{\beta }\right)}{\left(1+\cos 2\mathrm{\beta }\right)} \\ \Rightarrow \frac{2{\sin }^2\mathrm{\alpha }}{2{\cos }^2\mathrm{\alpha }}=\frac{2\left(2{\sin }^2\mathrm{\beta }\right)}{\left(2{\cos }^2\mathrm{\beta }\right)} \\ \Rightarrow {\tan }^2\mathrm{\alpha }=2{\tan }^2\mathrm{\beta } \\ \therefore \tan \mathrm{\alpha }=\sqrt{2}\tan \mathrm{\beta } \end{gathered}$$