In Delta ABC,ccos (A - alpha ) + acos (C + al | vedclass

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(B) We are given the expression $c\cos (A - \alpha ) + a\cos (C + \alpha )$.Using the trigonometric expansion formulas $\cos(x - y) = \cos x \cos y +

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$b\cos \alpha $

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(B) We are given the expression $c\cos (A - \alpha ) + a\cos (C + \alpha )$.Using the trigonometric expansion formulas $\cos(x - y) = \cos x \cos y + \sin x \sin y$ and $\cos(x + y) = \cos x \cos y - \sin x \sin y$,we get:$c(\cos A \cos \alpha + \sin A \sin \alpha) + a(\cos C \cos \alpha - \sin C \sin \alpha)$Rearranging the terms,we have:$\cos \alpha (c \cos A + a \cos C) + \sin \alpha (c \sin A - a \sin C)$From the projection rule in a triangle,$c \cos A + a \cos C = b$.From the sine rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,which implies $c \sin A = a \sin C$,so $c \sin A - a \sin C = 0$.Substituting these into the expression:$b \cos \alpha + \sin \alpha (0) = b \cos \alpha$.

In $\Delta ABC$,$c\cos (A - \alpha ) + a\cos (C + \alpha ) = $

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