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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]

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$\\\frac{\sin \left( x+y \right) }{\\sin \left( x - y \right) }=\frac{a+b}{a - b}~~~~ \\\\ ~~\frac{\\sin \left( x+y \right) +\\sin \left( x - y \right) }{\\sin \left( x - y \right) - \\sin \left( x - y \right) }=\frac{a+b+a - b}{a+b - a+b} \\\\$

$\\ \frac{2 \sin \frac{x+y+x - y}{2} \cos \frac{x+y - x+y}{2}}{2 \cos \frac{x+y+x - y}{2} \sin \frac{x+y - x+y}{2}} =\frac{2a}{2b} \\\\$

$\\\frac{\sin x\cos y}{\cos x\sin y}=\frac{a}{b} \\\\$

$\\ \frac{\tan x}{\tan y}=\frac{a}{b} \\\\$

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