Question

If $\frac{\sin (\alpha +\beta )}{\sin (\alpha -\beta )}=\frac{a+b}{a-b}$, where $\alpha \ne \beta ,a\ne b,b\ne 0$, then $\frac{\tan \alpha }{\tan \beta }$ is

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{a}{2b}$
• D. $-\frac{a}{b}$

Answer 1

The correct option is A $\frac{a}{b}$

$\frac{\sin (\alpha +\beta )}{\sin (\alpha -\beta )}=\frac{a+b}{a-b}\Rightarrow \frac{\sin \alpha \cos \beta +\sin \beta \cos \alpha }{\sin \alpha \cos \beta -\sin \beta \cos \alpha }=\frac{a+b}{a-b}$

Using componendo and divinendo,
$\Rightarrow \frac{\sin \alpha \cos \beta }{\sin \beta \cos \alpha }=\frac{a}{b}\Rightarrow \frac{\tan \alpha }{\tan \beta }=\frac{a}{b}$