If alpha + beta - gamma = pi, then sin^2 alph | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\alpha + \beta - \gamma = \pi$,which implies $\gamma = \alpha + \beta - \pi$.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^2 \beta - \sin^2 \gamma)$$= \sin^2 \alpha + \sin(\beta - \gamma) \sin(\beta + \gamma)$Since $\beta - \gamma = \pi - \alpha$,we have $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha$.Also,$\beta + \gamma = \beta + (\alpha + \beta - \pi) = \alpha + 2\beta - \pi$.Alternatively,using $\gamma = \alpha + \beta - \pi$,then $\sin \gamma = \sin(\alpha + \beta - \pi) = -\sin(\alpha + \beta)$.So,$\sin^2 \gamma = \sin^2(\alpha + \beta)$.Expression $= \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta)$$= \sin^2 \alpha + \sin^2 \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2$$= \sin^2 \alpha + \sin^2 \beta - (\sin^2 \alpha \cos^2 \beta + \cos^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta)$$= \sin^2 \alpha(1 - \cos^2 \beta) + \sin^2 \beta(1 - \cos^2 \alpha) - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$$= \sin^2 \alpha \sin^2 \beta + \sin^2 \beta \sin^2 \alpha - 2 \sin \alpha \sin \beta \cos \alpha \cos \beta$$= 2 \sin \alpha \sin \beta (\sin \alpha \sin \beta - \cos \alpha \cos \beta)$$= 2 \sin \alpha \sin \beta (- \cos(\alpha + \beta))$Since $\alpha + \beta = \pi + \gamma$,$\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma$.Therefore,the expression $= 2 \sin \alpha \sin \beta (-(- \cos \gamma)) = 2 \sin \alpha \sin \beta \cos \gamma$.

If $\alpha + \beta - \gamma = \pi,$ then $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

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