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(B) Let $y = \sin x$. The equation becomes $y^2 + by + c = 0$. Since $\alpha$ and $\beta$ are roots of the original equation,$\sin \alpha$ and $\sin \

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$2c$

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(B) Let $y = \sin x$. The equation becomes $y^2 + by + c = 0$. Since $\alpha$ and $\beta$ are roots of the original equation,$\sin \alpha$ and $\sin \beta$ are roots of the quadratic equation $y^2 + by + c = 0$. From the properties of quadratic equations,we have: $\sin \alpha + \sin \beta = -b$ $\sin \alpha \cdot \sin \beta = c$ Given $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$. Thus,$\sin \beta = \sin(\frac{\pi}{2} - \alpha) = \cos \alpha$. Substituting this into the product of roots: $\sin \alpha \cdot \cos \alpha = c$ $2 \sin \alpha \cos \alpha = 2c$ $\sin(2\alpha) = 2c$ Now,consider the sum of roots: $(\sin \alpha + \sin \beta)^2 = (-b)^2$ $\sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta = b^2$ Since $\sin \beta = \cos \alpha$,$\sin^2 \alpha + \cos^2 \alpha = 1$. $1 + 2c = b^2$ Therefore,$b^2 - 1 = 2c$.

$\alpha, \beta$ are the roots of the equation $\sin^2 x + b \sin x + c = 0$. If $\alpha + \beta = \frac{\pi}{2}$,then $b^2 - 1 =$

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