Question

$\text{If}\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b},$ then $\frac{\tan x}{\tan y}$ is equal to

• A. $\frac{b}{a}$
• B. $\frac{a}{b}$
• C. $ab$
• D. None of these

Answer 1

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

$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
$\Rightarrow \frac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}=\frac{a+b}{a-b}$
Applying componendo and dividendo, we get
$\frac{\tan x}{\tan y}=\frac{a}{b}$