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

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

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

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(D) Given: $\sin A + \sin B = C$ 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}$Now,using the double angle formula in terms of tangent:$\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(C/D)}{1 + (C/D)^2} = \frac{2C/D}{(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) = $

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