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If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then what is $\frac{\tan x}{\tan y}$ equal to?
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The correct answer is:
$\frac{a}{b}$
$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
Applying componendo and dividendo, we get
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cdot \cos y}{2 \cos x \cdot \sin y}=\frac{2 a}{2 b} \Rightarrow \tan x \cdot \cot y=\frac{a}{b}$
$\therefore \frac{\tan x}{\tan y}=\frac{a}{b}$
Applying componendo and dividendo, we get
$\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cdot \cos y}{2 \cos x \cdot \sin y}=\frac{2 a}{2 b} \Rightarrow \tan x \cdot \cot y=\frac{a}{b}$
$\therefore \frac{\tan x}{\tan y}=\frac{a}{b}$
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