If sin A + sin B = C and cos A + cos B = D,th | vedclass

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(D) Given that $\sin A + \sin B = C$ and $\cos A + \cos B = D$.Dividing the two equations:$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{C}{D}$Using

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$\frac{2CD}{C^2 + D^2}$

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(D) Given that $\sin A + \sin B = C$ and $\cos A + \cos B = D$.Dividing the two equations:$\frac{\sin A + \sin B}{\cos A + \cos B} = \frac{C}{D}$Using the sum-to-product formulas:$\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}} = \frac{C}{D}$$	an \frac{A+B}{2} = \frac{C}{D}$Using the identity $\sin 	heta = \frac{2 	an \frac{	heta}{2}}{1 + 	an^2 \frac{	heta}{2}}$ where $	heta = A+B$:$\sin (A+B) = \frac{2 	an \frac{A+B}{2}}{1 + 	an^2 \frac{A+B}{2}}$Substituting $	an \frac{A+B}{2} = \frac{C}{D}$:$\sin (A+B) = \frac{2(\frac{C}{D})}{1 + (\frac{C}{D})^2} = \frac{\frac{2C}{D}}{\frac{D^2 + C^2}{D^2}} = \frac{2C}{D} 	imes \frac{D^2}{C^2 + D^2} = \frac{2CD}{C^2 + D^2}$.

If $\sin A + \sin B = C$ and $\cos A + \cos B = D$,then the value of $\sin (A + B)$ is:

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