If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then show that $\frac{\tan x}{\tan y}=\frac{a}{b}$.
[Hint: Use componendo and Dividendo]
According to the question,
$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
Since, $\sin (A+B)=\sin A \cos B+\cos A \sin B$
$\therefore \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
$\Rightarrow \frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}=\frac{a+b}{a-b}$
Applying componendo-dividendo rule, We get,
$\Rightarrow \frac{(\sin x \cos y+\cos x \sin y)+(\sin x \cos y-\cos x \sin y)}{(\sin x \cos y+\cos x \sin y)-(\sin x \cos y-\cos x \sin y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cos y}{2 \cos x \sin y}=\frac{2 a}{2 b}$
$\Rightarrow\left(\frac{\sin x}{\cos x}\right)\left(\frac{\cos y}{\sin y}\right)=\frac{a}{b}$
Since, $\tan \mathrm{A}=(\sin \mathrm{A}) / \cos (\mathrm{A})$
$\Rightarrow \tan \mathrm{x}\left(\frac{1}{\tan \mathrm{y}}\right)=\frac{\mathrm{a}}{\mathrm{b}}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}$