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(D) Given,$	an A : 	an B : 	an C = 1 : 2 : 3$. We know that $	an A = \frac{\sin A}{\cos A} = \frac{a/2R}{(b^2+c^2-a^2)/2bc} = \frac{abc}{R(b^2+c^2

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$3$

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(D) Given,$	an A : 	an B : 	an C = 1 : 2 : 3$. We know that $	an A = \frac{\sin A}{\cos A} = \frac{a/2R}{(b^2+c^2-a^2)/2bc} = \frac{abc}{R(b^2+c^2-a^2)}$. Thus,$	an A : 	an B : 	an C = \frac{1}{b^2+c^2-a^2} : \frac{1}{a^2+c^2-b^2} : \frac{1}{a^2+b^2-c^2} = 1 : 2 : 3$. Let $b^2+c^2-a^2 = x$,$a^2+c^2-b^2 = 2x$,and $a^2+b^2-c^2 = 3x$. Adding these equations,we get $a^2+b^2+c^2 = 3x$. Then $a^2 = (a^2+b^2+c^2) - (b^2+c^2-a^2) = 3x - x = 2x$. $b^2 = (a^2+b^2+c^2) - (a^2+c^2-b^2) = 3x - 2x = x$. $c^2 = (a^2+b^2+c^2) - (a^2+b^2-c^2) = 3x - 3x = 0$. Wait,the ratio of sides $a:b:c$ is proportional to $\sin A : \sin B : \sin C$. Using the property $	an A : 	an B : 	an C = l : m : n$,we have $a : b : c = \sqrt{\frac{1}{m} + \frac{1}{n}} : \sqrt{\frac{1}{n} + \frac{1}{l}} : \sqrt{\frac{1}{l} + \frac{1}{m}}$. Substituting $l=1, m=2, n=3$: $a : b : c = \sqrt{\frac{1}{2} + \frac{1}{3}} : \sqrt{\frac{1}{3} + 1} : \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{5}{6}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{3}{2}} = \sqrt{\frac{5}{6}} : \sqrt{\frac{8}{6}} : \sqrt{\frac{9}{6}} = \sqrt{5} : 2\sqrt{2} : 3$. Since $\sin A : \sin B : \sin C = a : b : c$,we have $\sqrt{5} : 2\sqrt{2} : k = \sqrt{5} : 2\sqrt{2} : 3$. Therefore,$k = 3$.

In a $\triangle ABC$,if $\tan A : \tan B : \tan C = 1 : 2 : 3$ and $\sin A : \sin B : \sin C = \sqrt{5} : 2\sqrt{2} : k$,then $k =$

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