In a triangle ABC,with usual notations,frac{c | vedclass

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(D) Using the projection rule in $	riangle ABC$,we have $c = a \cos B + b \cos A$ and $b = a \cos C + c \cos A$.The given expression is $E = \frac{\c

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$\frac{1}{a}$

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(D) Using the projection rule in $	riangle ABC$,we have $c = a \cos B + b \cos A$ and $b = a \cos C + c \cos A$.The given expression is $E = \frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}$.Taking the common denominator: $E = \frac{a(\cos B+\cos C) + (b+c)\cos A}{a(b+c)}$.Expanding the numerator: $E = \frac{a \cos B + a \cos C + b \cos A + c \cos A}{a(b+c)}$.Rearranging terms: $E = \frac{(a \cos B + b \cos A) + (a \cos C + c \cos A)}{a(b+c)}$.Substituting the projection rules: $E = \frac{c + b}{a(b+c)}$.Simplifying: $E = \frac{b+c}{a(b+c)} = \frac{1}{a}$.

In a triangle $ABC$,with usual notations,$\frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}$ has the value

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