If sin x - sin y = frac{1}{2} and cos x - cos | vedclass

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(C) Given,$\sin x - \sin y = \frac{1}{2} \dots (i)$ and $\cos x - \cos y = 1 \dots (ii)$.Using sum-to-product formulas:$2 \cos \left(\frac{x+y}{2}\rig

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$\frac{4}{3}$

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(C) Given,$\sin x - \sin y = \frac{1}{2} \dots (i)$ and $\cos x - \cos y = 1 \dots (ii)$.Using sum-to-product formulas:$2 \cos \left(\frac{x+y}{2}ight) \sin \left(\frac{x-y}{2}ight) = \frac{1}{2} \dots (iii)$$-2 \sin \left(\frac{x+y}{2}ight) \sin \left(\frac{x-y}{2}ight) = 1 \dots (iv)$Dividing $(iv)$ by $(iii)$:$\frac{-2 \sin \left(\frac{x+y}{2}ight) \sin \left(\frac{x-y}{2}ight)}{2 \cos \left(\frac{x+y}{2}ight) \sin \left(\frac{x-y}{2}ight)} = \frac{1}{1/2}$$-	an \left(\frac{x+y}{2}ight) = 2 \implies 	an \left(\frac{x+y}{2}ight) = -2$.Using the formula $	an(2	heta) = \frac{2	an 	heta}{1 - 	an^2 	heta}$ where $	heta = \frac{x+y}{2}$:$	an(x+y) = \frac{2 	an \left(\frac{x+y}{2}ight)}{1 - 	an^2 \left(\frac{x+y}{2}ight)} = \frac{2(-2)}{1 - (-2)^2} = \frac{-4}{1 - 4} = \frac{-4}{-3} = \frac{4}{3}$.

If $\sin x - \sin y = \frac{1}{2}$ and $\cos x - \cos y = 1$,then $\tan(x + y)$ is equal to

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