यदि alpha + beta - gamma = pi है,तो sin^2 alp | vedclass

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Question

(A) दिया गया है $\alpha + \beta - \gamma = \pi$,जिसका अर्थ है $\gamma = \alpha + \beta - \pi$.सर्वसमिका $\sin^2 A - \sin^2 B = \sin(A - B) \sin(A + B)

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

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(A) दिया गया है $\alpha + \beta - \gamma = \pi$,जिसका अर्थ है $\gamma = \alpha + \beta - \pi$.सर्वसमिका $\sin^2 A - \sin^2 B = \sin(A - B) \sin(A + B)$ का उपयोग करने पर:$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = \sin^2 \alpha + (\sin^2 \beta - \sin^2 \gamma)$$= \sin^2 \alpha + \sin(\beta - \gamma) \sin(\beta + \gamma)$चूंकि $\beta - \gamma = \pi - \alpha$,इसलिए $\sin(\beta - \gamma) = \sin(\pi - \alpha) = \sin \alpha$.साथ ही,$\gamma = \alpha + \beta - \pi$ होने के कारण,$\sin \gamma = \sin(\alpha + \beta - \pi) = -\sin(\alpha + \beta)$.अतः,$\sin^2 \gamma = \sin^2(\alpha + \beta)$.व्यंजक $= \sin^2 \alpha + \sin^2 \beta - \sin^2(\alpha + \beta)$$= \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))$चूंकि $\alpha + \beta = \pi + \gamma$,इसलिए $\cos(\alpha + \beta) = \cos(\pi + \gamma) = -\cos \gamma$.अतः,व्यंजक $= 2 \sin \alpha \sin \beta (-(- \cos \gamma)) = 2 \sin \alpha \sin \beta \cos \gamma$.

यदि $\alpha + \beta - \gamma = \pi$ है,तो $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

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