Question

If $\cos 2\alpha =\frac{(3\cos 2\beta -1)}{(3-\cos 2\beta )}$, then $\tan \alpha$is equal to

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

Answer 1

The correct option is A

$\sqrt{2}\tan \beta$

Explanation for the correct option:
Finding the value of $\tan \alpha$:

Given

$$\begin{gathered} \Rightarrow \cos 2\alpha =\frac{(3\cos 2\beta –1)}{(3–\cos 2\beta )}\dots .\left(i\right) \\ \Rightarrow {\tan }^2\alpha =\frac{\left({\sin }^2\alpha \right)}{\left({\cos }^2\alpha \right)} \\ =\frac{\left(2{\sin }^2\alpha \right)}{\left(2{\cos }^2\alpha \right)} \\ =\frac{(1–\cos 2\alpha )}{(1+\cos 2\alpha )} \end{gathered}$$

Substituting in equation $\left(1\right)$

$$\begin{gathered} ta{n}^2\alpha =\frac{[1–{\displaystyle \frac{(3\cos 2\beta –1)}{(3–\cos 2\beta }}]}{[1+\frac{(3\cos 2\beta –1)}{(3–\cos 2\beta }]} \\ =\frac{(3–\cos 2\beta –3\cos 2\beta +1)}{(3–cos2\beta +3cos2\beta –1)} \\ =\frac{(4–4\cos 2\beta )}{(2+2cos2\beta )} \\ =\frac{2(1–\cos 2\beta )}{(1+cos2\beta )} \\ =\frac{2\left(2{\sin }^2\beta \right)}{\left(2co{s}^2\beta \right)} \\ =2{\text{tan}}^2\beta \end{gathered}$$

$\mathbf{\therefore }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{2}}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\beta }$

Hence, Option ‘A’ is Correct.