Question

If $\tan \alpha =\frac{1}{7},\tan \beta =\frac{1}{3}$, then $\cos 2\alpha$ is equal to

(a) $\sin 2\beta$ (b) $\sin 4\beta$ (c) $\sin 3\beta$ (d) $\cos 2\beta$

Answer 1

It is given that $\tan \alpha =\frac{1}{7}$ and $\tan \beta =\frac{1}{3}$.

Now,

$$\begin{gathered} \tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta } \\ =\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{9}}} \\ =\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{8}{9}}} \\ =\frac{3}{4} \end{gathered}$$

$$\begin{gathered} \therefore \tan \left(\alpha +2\beta \right)=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta } \\ =\frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{1}{7}}\times {\displaystyle \frac{3}{4}}} \\ =\frac{{\displaystyle \frac{25}{28}}}{{\displaystyle \frac{25}{28}}} \\ =1 \end{gathered}$$

$$\begin{gathered} \tan \left(\alpha +2\beta \right)=1=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \alpha +2\beta =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4}-2\beta \\ \Rightarrow 2\alpha =\frac{\mathrm{\pi }}{2}-4\beta \\ \Rightarrow \cos 2\alpha =\cos \left(\frac{\mathrm{\pi }}{2}-4\beta \right)=\sin 4\beta \end{gathered}$$

$\therefore \cos 2\alpha =\sin 4\beta$

Hence, the correct answer is option B.