In triangle ABC,if tan frac{A}{2}+tan frac{C} | vedclass

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(D) Given $	an \frac{A}{2}+	an \frac{C}{2}=\frac{b}{s}$.Using the identity $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,we have:$\frac{\sin

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$\frac{1}{2}$

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(D) Given $	an \frac{A}{2}+	an \frac{C}{2}=\frac{b}{s}$.Using the identity $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,we have:$\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} + \frac{\sin \frac{C}{2}}{\cos \frac{C}{2}} = \frac{\sin \frac{A+C}{2}}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{b}{s}$.Substituting $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$ and $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we get:$\frac{\sin \frac{A+C}{2}}{\sqrt{\frac{s^2(s-a)(s-c)}{ab^2c}}} = \frac{b}{s}$ $\Rightarrow \frac{\sin \frac{A+C}{2}}{\frac{s}{b} \sqrt{\frac{(s-a)(s-c)}{ac}}} = \frac{b}{s}$.Since $\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$,the equation becomes $\frac{\sin \frac{A+C}{2}}{\sin \frac{B}{2}} = 1$.Since $A+B+C = \pi$,$\frac{B}{2} = \frac{\pi}{2} - \frac{A+C}{2}$,so $\sin \frac{B}{2} = \cos \frac{A+C}{2}$.Thus,$	an \frac{A+C}{2} = 1$,which implies $\frac{A+C}{2} = \frac{\pi}{4}$,so $A+C = \frac{\pi}{2}$.Finally,$\sin \left(\frac{A+C}{3}ight) = \sin \left(\frac{\pi/2}{3}ight) = \sin \left(\frac{\pi}{6}ight) = \frac{1}{2}$.

In $\triangle ABC$,if $\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{b}{s}$,then $\sin \left(\frac{A+C}{3}\right)=$

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