If x neq -y and sin x + sin y = 3(cos y - cos | vedclass

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(C) Given,$\sin x + \sin y = 3(\cos y - \cos x)$ Using sum-to-product formulas: $2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = 3

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) Given,$\sin x + \sin y = 3(\cos y - \cos x)$ Using sum-to-product formulas: $2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = 3 \left( -2 \sin \left(\frac{y+x}{2}ight) \sin \left(\frac{y-x}{2}ight) ight)$ Since $x 
eq -y$,$\sin \left(\frac{x+y}{2}ight) 
eq 0$,we can divide both sides by $2 \sin \left(\frac{x+y}{2}ight)$: $\cos \left(\frac{x-y}{2}ight) = -3 \sin \left(\frac{y-x}{2}ight) = 3 \sin \left(\frac{x-y}{2}ight)$ $	an \left(\frac{x-y}{2}ight) = \frac{1}{3}$ Now,using the double angle formula $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta}$: $	an(x-y) = \frac{2 	an \left(\frac{x-y}{2}ight)}{1 - 	an^2 \left(\frac{x-y}{2}ight)}$ $	an(x-y) = \frac{2 	imes \frac{1}{3}}{1 - (\frac{1}{3})^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$

If $x \neq -y$ and $\sin x + \sin y = 3(\cos y - \cos x)$,then $\tan(x - y) =$

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