If frac{sin(x+y)}{sin(x-y)} = frac{a+b}{a-b}, | vedclass

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(B) Given,$\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$ Applying Componendo and Dividendo: $\frac{\sin(x+y) + \sin(x-y)}{\sin(x+y) - \sin(x-y)} = \f

Vedclass · Accepted Answer

$\frac{a}{b}$

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(B) Given,$\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$ Applying Componendo and Dividendo: $\frac{\sin(x+y) + \sin(x-y)}{\sin(x+y) - \sin(x-y)} = \frac{(a+b) + (a-b)}{(a+b) - (a-b)}$ Using the expansion formulas $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$: $\frac{2 \sin x \cos y}{2 \cos x \sin y} = \frac{2a}{2b}$ $\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$ $\frac{	an x}{	an y} = \frac{a}{b}$

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

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