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(D) Given that $\alpha$ and $\beta$ are roots of $\sin^2 x + a \sin x + b = 0$,we have $\sin \alpha + \sin \beta = -a$.Given that $\alpha$ and $\beta$

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$\frac{2ac}{a^2 + c^2}$

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(D) Given that $\alpha$ and $\beta$ are roots of $\sin^2 x + a \sin x + b = 0$,we have $\sin \alpha + \sin \beta = -a$.Given that $\alpha$ and $\beta$ are roots of $\cos^2 x + c \cos x + d = 0$,we have $\cos \alpha + \cos \beta = -c$.Using sum-to-product formulas:$2 \sin \left( \frac{\alpha + \beta}{2} ight) \cos \left( \frac{\alpha - \beta}{2} ight) = -a$  $(1)$$2 \cos \left( \frac{\alpha + \beta}{2} ight) \cos \left( \frac{\alpha - \beta}{2} ight) = -c$  $(2)$Dividing $(1)$ by $(2)$:$	an \left( \frac{\alpha + \beta}{2} ight) = \frac{a}{c}$Using the identity $\sin 	heta = \frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)}$:$\sin(\alpha + \beta) = \frac{2 	an \left( \frac{\alpha + \beta}{2} ight)}{1 + 	an^2 \left( \frac{\alpha + \beta}{2} ight)} = \frac{2(a/c)}{1 + (a/c)^2} = \frac{2ac}{a^2 + c^2}$.

If $\alpha$ and $\beta$ are solutions of $\sin^2 x + a \sin x + b = 0$ as well as $\cos^2 x + c \cos x + d = 0$,then $\sin(\alpha + \beta)$ is equal to

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