If alpha+beta+gamma=pi,then the expression si | vedclass

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(D) Given $\alpha+\beta+\gamma=\pi$,so $\gamma = \pi - (\alpha+\beta)$.Using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:$\sin^2 \alpha +

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$2 \sin \alpha \sin \beta \cos \gamma$

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(D) Given $\alpha+\beta+\gamma=\pi$,so $\gamma = \pi - (\alpha+\beta)$.Using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + \sin(\beta+\gamma)\sin(\beta-\gamma)$Since $\beta+\gamma = \pi - \alpha$,$\sin(\beta+\gamma) = \sin(\pi-\alpha) = \sin \alpha$.Substituting this:$= \sin^2 \alpha + \sin \alpha \sin(\beta-\gamma)$$= \sin \alpha [\sin \alpha + \sin(\beta-\gamma)]$$= \sin \alpha [\sin(\beta+\gamma) + \sin(\beta-\gamma)]$Using the sum-to-product formula $\sin(x+y) + \sin(x-y) = 2 \sin x \cos y$:$= \sin \alpha [2 \sin \beta \cos \gamma]$$= 2 \sin \alpha \sin \beta \cos \gamma$.

If $\alpha+\beta+\gamma=\pi$,then the expression $\sin^2 \alpha+\sin^2 \beta-\sin^2 \gamma$ is equal to:

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