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(D) Given the expression $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.Simplifying the terms inside the parentheses:$

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$3$

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(D) Given the expression $\left(1+\frac{a}{c}+\frac{b}{c}ight)\left(1+\frac{c}{b}-\frac{a}{b}ight)$.Simplifying the terms inside the parentheses:$= \left(\frac{c+a+b}{c}ight) \left(\frac{b+c-a}{b}ight)$$= \frac{(b+c)+a}{c} 	imes \frac{(b+c)-a}{b}$$= \frac{(b+c)^2 - a^2}{bc}$$= \frac{b^2 + c^2 + 2bc - a^2}{bc}$$= \frac{b^2 + c^2 - a^2}{bc} + 2$Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,we have $\frac{b^2 + c^2 - a^2}{bc} = 2 \cos A$.$= 2 \cos A + 2$Given $\angle A = 60^{\circ}$,$\cos 60^{\circ} = 1/2$.$= 2(1/2) + 2 = 1 + 2 = 3$.

In a triangle $ABC$,with usual notations $\angle A=60^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=$

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