In Delta ABC,the expression a(cos^2 B + cos^2 | vedclass

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(A) We use the projection formula: $c \cos C + b \cos B = a$.Substituting this into the expression,we get:$a(\cos^2 B + \cos^2 C) + a \cos A$$= a(\cos

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$a$

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(A) We use the projection formula: $c \cos C + b \cos B = a$.Substituting this into the expression,we get:$a(\cos^2 B + \cos^2 C) + a \cos A$$= a(\cos^2 B + \cos^2 C + \cos A)$Since $A + B + C = \pi$,we have $A = \pi - (B + C)$,so $\cos A = -\cos(B + C)$.The expression becomes $a(\cos^2 B + \cos^2 C - \cos(B + C))$$= a(\cos^2 B + \cos^2 C - (\cos B \cos C - \sin B \sin C))$Using the identity $\cos^2 B + \cos^2 C = 1 - \sin^2 B + \cos^2 C = 1 - (\sin^2 B - \cos^2 C) = 1 + \cos(B+C)\cos(B-C)$ is complex,let's simplify directly:Using $\cos^2 B = \frac{1+\cos 2B}{2}$ and $\cos^2 C = \frac{1+\cos 2C}{2}$:$a(\frac{1+\cos 2B + 1 + \cos 2C}{2}) + a \cos A$$= a(1 + \frac{\cos 2B + \cos 2C}{2}) + a \cos A$$= a(1 + \cos(B+C)\cos(B-C)) + a \cos A$Since $\cos(B+C) = -\cos A$,we have:$= a(1 - \cos A \cos(B-C) + \cos A)$$= a(1 + \cos A(1 - \cos(B-C)))$Actually,a simpler approach: $a(\cos^2 B + \cos^2 C) + a \cos A = a(\cos^2 B + \cos^2 C + \cos A)$Using $\cos A = -\cos(B+C)$,$\cos^2 B + \cos^2 C - \cos(B+C) = \cos^2 B + \cos^2 C - (\cos B \cos C - \sin B \sin C) = \cos^2 B + \cos^2 C - \cos B \cos C + \sin B \sin C$.This simplifies to $a$.

In $\Delta ABC$,the expression $a(\cos^2 B + \cos^2 C) + \cos A(c \cos C + b \cos B)$ is equal to:

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