In a triangle ABC, if $\tan A : \tan B : \tan C =1 : 2 : 3$, then $a^2 : b^{2 } : c^2 =$

Solution:

$\frac{\tan A}{1}=\frac{\tan B}{2}=\frac{\tan C}{3}=k$ (say)

$\in \tan A=\pi \tan A \Rightarrow 6k=6 k^3$

$\Rightarrow k=1$

$\therefore \tan A=1, \tan B=2, \tan C=3$

$\Rightarrow cotA=1, cotB=\frac{1}{2}, cotC=\frac{1}{3},$

 $\begin{array}{l}\Rightarrow {\mathrm{cosec}}^{2}A=2,{\mathrm{cosec}}^{2}B=\frac{5}{4},{\mathrm{cosec}}^{2}C=\frac{10}{9}\\ \Rightarrow {\sin }^{2}A=\frac{1}{2},{\sin }^{2}B=\frac{4}{5},{\sin }^{2}C=\frac{9}{10}\\ \therefore a^2:b^2:c^2={\sin }^{2}A:{\sin }^{2}B:{\sin }^{2}C\\ =\frac{1}{2}:\frac{4}{5}:\frac{9}{10}=5:8:9.\end{array}$