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

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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 $\sin \alpha + \sin \beta = -a$.Similarly,for $\cos^2 x + c \cos x + d = 0$,we have $\cos \alpha + \cos \beta = -c$.Using sum-to-product formulas:$2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -a$  .....$(1)$$2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} = -c$  .....$(2)$Dividing equation $(1)$ by $(2)$:$\frac{\sin \frac{\alpha + \beta}{2}}{\cos \frac{\alpha + \beta}{2}} = \frac{-a}{-c} \Rightarrow 	an \frac{\alpha + \beta}{2} = \frac{a}{c}$.Using the identity $\sin 	heta = \frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)}$ where $	heta = \alpha + \beta$:$\sin(\alpha + \beta) = \frac{2(a/c)}{1 + (a/c)^2} = \frac{2a/c}{(c^2 + a^2)/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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