If alpha, beta, gamma are angles of a triangl | vedclass

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(A) Given that $\alpha + \beta + \gamma = \pi$,so $\gamma = \pi - (\alpha + \beta)$.Then $\cos \gamma = -\cos(\alpha + \beta)$.The expression is $E =

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$2$

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(A) Given that $\alpha + \beta + \gamma = \pi$,so $\gamma = \pi - (\alpha + \beta)$.Then $\cos \gamma = -\cos(\alpha + \beta)$.The expression is $E = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$E = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$$E = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$.Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 - 2 \cos \alpha \cos \beta \cos \gamma$,we substitute this:$E = 3 - (1 - 2 \cos \alpha \cos \beta \cos \gamma) - 2 \cos \alpha \cos \beta \cos \gamma$$E = 3 - 1 + 2 \cos \alpha \cos \beta \cos \gamma - 2 \cos \alpha \cos \beta \cos \gamma$$E = 2$.

If $\alpha, \beta, \gamma$ are angles of a triangle,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma - 2 \cos \alpha \cos \beta \cos \gamma$ is

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