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

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(D) Given equations are: $\sin x + \sin y = \frac{1}{2} \quad (1)$ $\cos x + \cos y = 1 \quad (2)$ Using sum-to-product formulas: $2 \sin \left(\frac{

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$\frac{4}{3}$

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(D) Given equations are: $\sin x + \sin y = \frac{1}{2} \quad (1)$ $\cos x + \cos y = 1 \quad (2)$ Using sum-to-product formulas: $2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{1}{2} \quad (3)$ $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = 1 \quad (4)$ Dividing equation $(3)$ by $(4)$: $\frac{2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)}{2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)} = \frac{1/2}{1}$ $	an \left(\frac{x+y}{2}ight) = \frac{1}{2}$ Using the double angle 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)}$ Substituting $	an \left(\frac{x+y}{2}ight) = \frac{1}{2}$: $	an(x+y) = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$

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

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