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(D) Given that $\Delta$ is the area of triangle $ABC$. Using the projection rule,we have $b \cos C + c \cos B = a$. Using the sine rule,we have $\sin

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$4 \Delta$

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(D) Given that $\Delta$ is the area of triangle $ABC$. Using the projection rule,we have $b \cos C + c \cos B = a$. Using the sine rule,we have $\sin C = \frac{c}{2R}$ and $\sin B = \frac{b}{2R}$,where $R$ is the circumradius. Thus,$b \sin C + c \sin B = b(\frac{c}{2R}) + c(\frac{b}{2R}) = \frac{bc}{2R} + \frac{cb}{2R} = \frac{2bc}{2R} = \frac{bc}{R}$. Now,the expression becomes: $(b \sin C + c \sin B)(b \cos C + c \cos B) = (\frac{bc}{R})(a) = \frac{abc}{R}$. Since the area of a triangle is given by $\Delta = \frac{abc}{4R}$,we have $\frac{abc}{R} = 4 \Delta$. Therefore,the correct option is $4 \Delta$.

If $\Delta$ denotes the area of triangle $ABC$,then $(b \sin C + c \sin B)(b \cos C + c \cos B) =$

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