MCQ
If $5\sin\alpha=3\sin(\alpha+2\beta)\not=0,$ then $\tan(\alpha+\beta)$ is equal to:
- A$2\tan\beta$
- B$3\tan\beta$
- ✓$4\tan\beta$
- D$6\tan\beta$
✓
Answer
Correct option: C.
$4\tan\beta$
We have,
$5\sin\alpha=3\sin(\alpha+2\beta)$
$\Rightarrow\frac{5}{3}=\frac{\sin(\alpha+2\beta)}{\sin\alpha}$
$\Rightarrow\frac{5-3}{5+3}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$ (using componendo and dividendo)
$\Rightarrow\frac{2}{8}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$
$\Rightarrow\frac{1}{4}=\frac{2\cos\frac{\alpha+2\beta+\alpha}{2}\sin\frac{\alpha+2\beta-\alpha}{2}}{2\sin\frac{\alpha+2\beta+\alpha}{2}\cos\frac{\alpha+2\beta-\alpha}{2}}$
$\Rightarrow\frac{1}{4}\frac{\cos(\alpha+\beta)\sin\beta}{\sin(\alpha+\beta)\cos\beta}$
$\Rightarrow\frac{1}{4}=\cot(\alpha+\beta)\tan\beta$
$\Rightarrow\frac{1}{4}=\frac{1}{\tan(\alpha+\beta)}\tan\beta$
$\therefore\tan(\alpha+\beta)=4\tan\beta$
$5\sin\alpha=3\sin(\alpha+2\beta)$
$\Rightarrow\frac{5}{3}=\frac{\sin(\alpha+2\beta)}{\sin\alpha}$
$\Rightarrow\frac{5-3}{5+3}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$ (using componendo and dividendo)
$\Rightarrow\frac{2}{8}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$
$\Rightarrow\frac{1}{4}=\frac{2\cos\frac{\alpha+2\beta+\alpha}{2}\sin\frac{\alpha+2\beta-\alpha}{2}}{2\sin\frac{\alpha+2\beta+\alpha}{2}\cos\frac{\alpha+2\beta-\alpha}{2}}$
$\Rightarrow\frac{1}{4}\frac{\cos(\alpha+\beta)\sin\beta}{\sin(\alpha+\beta)\cos\beta}$
$\Rightarrow\frac{1}{4}=\cot(\alpha+\beta)\tan\beta$
$\Rightarrow\frac{1}{4}=\frac{1}{\tan(\alpha+\beta)}\tan\beta$
$\therefore\tan(\alpha+\beta)=4\tan\beta$
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