若$\Delta ABC$满足:$\tan\dfrac{A}{2}\cdot\tan\dfrac{C}{2}=\dfrac{1}{3},b=\dfrac{4}{3}a$,则$\sin B=$______
分析:$\tan\dfrac{A}{2}\cdot\tan\dfrac{C}{2}=\dfrac{1-\cos A}{\sin A}\cdot\dfrac{\sin C}{1+\cos C}=\dfrac{1}{3}$
得 $\dfrac{a^2-(b-c)^2}{(a+b)^2-c^2}=\dfrac{1}{3}$,得$\dfrac{a-b+c}{a+b+c}=\dfrac{1}{3}$,得$a+c=2b,c=\dfrac{5}{3}a$
故$\angle C=\dfrac{\pi}{2},\sin B =\dfrac{b}{c}=\dfrac{4}{5}$
另附上我两个同事的解答：
注:$\sum\limits_{cyc}tan\dfrac{A}{2}\tan\dfrac{B}{2}=1$