In triangle ABC,if theta is any angle,then b | vedclass

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(B) We are given the expression $b \cos (C+	heta) + c \cos (B-	heta)$.Expanding the cosine terms using the identity $\cos(x \pm y) = \cos x \cos y \

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$a \cos 	heta$

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(B) We are given the expression $b \cos (C+	heta) + c \cos (B-	heta)$.Expanding the cosine terms using the identity $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$:$= b(\cos C \cos 	heta - \sin C \sin 	heta) + c(\cos B \cos 	heta + \sin B \sin 	heta)$$= (b \cos C + c \cos B) \cos 	heta - (b \sin C - c \sin B) \sin 	heta$Using the projection formula,$b \cos C + c \cos B = a$.Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $b \sin C = c \sin B$,so $b \sin C - c \sin B = 0$.Substituting these into the expression:$= a \cos 	heta - 0 \cdot \sin 	heta = a \cos 	heta$.

In $\triangle ABC$,if $\theta$ is any angle,then $b \cos (C+\theta) + c \cos (B-\theta) =$

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