Question

If $\sin (\alpha +\beta )=6\sin (\alpha -\beta )$ and $k=\frac{\tan \alpha }{\tan \beta }$, then, what is the value of $k$ ?

• (A) $\frac{3}{5}$
• (B) $\frac{6}{5}$
• (C) $\frac{4}{5}$
• (D) $\frac{7}{5}$

Answer 1

The Correct Option is D

Explanation:
Given,$\sin (\alpha +\beta )=6\sin (\alpha -\beta )\text{ and }k=\frac{\tan \alpha }{\tan \beta }$$\sin (\alpha +\beta )=6\sin (\alpha -\beta )$$\Rightarrow \sin \alpha \cos \beta +\cos \alpha \sin \beta =6\sin \alpha \cos \beta -6\cos \alpha \sin \beta$$\Rightarrow \cos \alpha \sin \beta +6\cos \alpha \sin \beta =6\sin \alpha \cos \beta -\sin \alpha \cos \beta$$\Rightarrow 7\cos \alpha \sin \beta =5\sin \alpha \cos \beta$$\Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{7}{5}\times \frac{\sin \beta }{\cos \beta }$$\Rightarrow \text{ tan }\alpha =\frac{7}{5}\times \tan \beta$$\Rightarrow k=\frac{\tan \alpha }{\tan \beta }=\frac{7}{5}$$\therefore$ The value of $k\text{ is }\frac{7}{5}$
Hence, the correct option is (D).