If alpha, beta, gamma in left( 0, frac{pi}{2} | vedclass

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(A) We have $\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma)$.Using the expansion $\sin(\alpha + \beta + \gamma) = \sin \alpha

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(A) We have $\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma)$.Using the expansion $\sin(\alpha + \beta + \gamma) = \sin \alpha \cos \beta \cos \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma - \sin \alpha \sin \beta \sin \gamma$.Substituting this into the expression:$= \sin \alpha(1 - \cos \beta \cos \gamma) + \sin \beta(1 - \cos \alpha \cos \gamma) + \sin \gamma(1 - \cos \alpha \cos \beta) + \sin \alpha \sin \beta \sin \gamma$.Since $\alpha, \beta, \gamma \in (0, \pi/2)$,we have $\cos \beta \cos \gamma Thus,each term is positive,so $\sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma) > 0$.Therefore,$\sin \alpha + \sin \beta + \sin \gamma > \sin(\alpha + \beta + \gamma)$.This implies $\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma} < 1$.

If $\alpha, \beta, \gamma \in \left( 0, \frac{\pi}{2} \right)$,then $\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma}$ is

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