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

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(B) Given: $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$Applying the Componendo and Dividendo rule,which states that if $\frac{p}{q} = \frac

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$\frac{a}{b}$

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(B) Given: $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$Applying the Componendo and Dividendo rule,which states that if $\frac{p}{q} = \frac{r}{s},$ then $\frac{p + q}{p - q} = \frac{r + s}{r - s}$:$\frac{\sin(x + y) + \sin(x - y)}{\sin(x + y) - \sin(x - y)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)}$Using the trigonometric identities $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$ and $\sin(A + B) - \sin(A - B) = 2\cos A \sin B$:$\frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}$$\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$$	an x \cdot \frac{1}{	an 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 $\frac{\tan x}{\tan y}$ is equal to

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