cos alpha sin (beta - gamma ) + cos beta sin | vedclass

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(A) We have the expression: $E = \cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta )$.Expandi

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We have the expression: $E = \cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta )$.Expanding each term using the identity $\sin(x - y) = \sin x \cos y - \cos x \sin y$:$E = \cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.Distributing the terms:$E = \cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta$.Observing the terms:$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$.Thus,the correct option is $A$.

$\cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta ) = $

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