Trigonometry! 20

We have: $\Large{\frac{2\tan A}{1-\tan^2A}=\frac{12}{5}}$

$\Large{\Rightarrow \frac{12 \tan^2A+10 \tan A-12}{5-5 \tan^2A}=0}$

$\Large{\Rightarrow \frac{2(2 \tan A+3)(3 \tan A-2)}{5-5 \tan^2A}=0}$

Since we have $0<A<\frac{\pi}{2}$ , we have $\tan A>0$ and hence $\tan A=\frac{2}{3}$

Then we have: $70+9\tan3A=70+9 \cdot \Large{\frac{\frac{2}{3}+\frac{12}{5}}{1-\frac{2}{3}\cdot\frac{12}{5}}}$ $=70+9\cdot \Large{\frac{\frac{46}{15}}{1-\frac{24}{15}}}$

$=70+9\cdot\frac{46}{15-24}=70-46=\boxed{24}$

Can someone suggest a method using which we can reach tan3A directly from tan2A ,without taking out tanA ?