Question

If $5\sin \alpha =3\sin \left(\alpha +2\beta \right)\ne 0$, then $\tan \left(\alpha +\beta \right)$ is equal to
(a) $2\tan \beta$
(b) $3\tan \beta$
(c) $4\tan \beta$
(d) $6\tan \beta$

Answer 1

(c) $4\tan \beta$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ 5\sin \alpha =3\sin \left(\alpha +2\beta \right) \\ \Rightarrow \frac{5}{3}=\frac{\sin \left(\alpha +2\beta \right)}{\sin \alpha } \\ \Rightarrow \frac{5-3}{5+3}=\frac{\sin \left(\alpha +2\beta \right)-\sin \alpha }{\sin \left(\alpha +2\beta \right)+\sin \alpha }\left(\mathrm{Using}\mathrm{componendo}\mathrm{and}\mathrm{d}\mathrm{ividendo}\right) \\ \Rightarrow \frac{2}{8}=\frac{\sin \left(\alpha +2\beta \right)-\sin \alpha }{\sin \left(\alpha +2\beta \right)+\sin \alpha } \\ \Rightarrow \frac{1}{4}=\frac{2\cos {\displaystyle \frac{\alpha +2\beta +\alpha }{2}}\sin {\displaystyle \frac{\alpha +2\beta -\alpha }{2}}}{2\sin {\displaystyle \frac{\alpha +2\beta +\alpha }{2}}\cos {\displaystyle \frac{\alpha +2\beta -\alpha }{2}}} \\ \Rightarrow \frac{1}{4}=\frac{{\displaystyle \cos \left(\alpha +\beta \right)\sin \beta }}{\sin \left(\alpha +\beta \right)\cos \beta } \\ \Rightarrow \frac{1}{4}=\cot \left(\alpha +\beta \right)\tan \beta \\ \Rightarrow \frac{1}{4}=\frac{1}{\tan \left(\alpha +\beta \right)}\tan \beta \\ \therefore \tan \left(\alpha +\beta \right)=4\tan \beta \end{gathered}$$