In triangle ABC,if B+C=72^{circ},then left(1+ | vedclass

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(D) Given expression: $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right) = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b

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$\frac{5-\sqrt{5}}{2}$

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(D) Given expression: $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right) = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$$= \frac{(b+c)^2-a^2}{bc} = \frac{b^2+c^2-a^2+2bc}{bc}$$= \frac{b^2+c^2-a^2}{bc} + 2 = 2\left(\frac{b^2+c^2-a^2}{2bc}\right) + 2$$= 2\cos A + 2$Since $A+B+C = 180^{\circ}$ and $B+C = 72^{\circ}$,we have $A = 180^{\circ} - 72^{\circ} = 108^{\circ}$.So,the expression becomes $2\cos 108^{\circ} + 2 = 2\cos(90^{\circ} + 18^{\circ}) + 2$$= -2\sin 18^{\circ} + 2 = -2\left(\frac{\sqrt{5}-1}{4}\right) + 2$$= \frac{1-\sqrt{5}}{2} + 2 = \frac{1-\sqrt{5}+4}{2} = \frac{5-\sqrt{5}}{2}$.

In $\triangle ABC$,if $B+C=72^{\circ}$,then $\left(1+\frac{a}{c}+\frac{b}{c}\right) \left(1+\frac{c}{b}-\frac{a}{b}\right)$ is equal to

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