2)
$\frac{12tan^{2}A-4tan^{4}A}{6tanA-18tan^{3}A} =36cos^{2}10^{\circ }-96sin^{4}80^{\circ }+64cos^{6}370^{\circ }$
$\frac{(4tanA)(3tanA-tan^{3}A)}{(6tanA)(1-3tan^{2}A)} =36cos^{2}10^{\circ }-96cos^{4}10^{\circ }+64cos^{6}10^{\circ }$
$\left(\frac{2}{3} \right) (tan3A)=18(2cos^{2}10^{\circ })-24(2cos^{2}10^{\circ })^{2}+8(2cos^{2}10^{\circ })^{3}$
$\left(\frac{2}{3} \right) (tan3A)=18(cos20^{\circ }+1)-24(cos20^{\circ }+1)^{2}+8(cos20^{\circ }+1)^{3}$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(18-24(cos20^{\circ }+1)+8(cos20^{\circ }+1)^{2})$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(18-(cos20^{\circ }+1)(24-8(cos20^{\circ }+1)))$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(18-(cos20^{\circ }+1)(16-8cos20^{\circ }))$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(18-16-8cos20^{\circ }+8cos^{2}20^{\circ })$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(2-8cos20^{\circ }+4(cos40^{\circ }+1))$
$\left(\frac{2}{3} \right) (tan3A)=(cos20^{\circ }+1)(6-8cos20^{\circ }+4cos40^{\circ })$
$\left(\frac{2}{3} \right) (tan3A)=6cos20^{\circ }-8cos^{2}20^{\circ }+4cos40^{\circ }cos20^{\circ }+6-8cos20^{\circ }+4cos40^{\circ }$
$\left(\frac{2}{3} \right) (tan3A)=6cos20^{\circ }-4(cos40^{\circ }+1)+2(cos60^{\circ }+cos20^{\circ })+6-8cos20^{\circ }+4cos40^{\circ }$
$\left(\frac{2}{3} \right) (tan3A)=3+8cos20^{\circ }-4cos40^{\circ }-8cos20^{\circ }+4cos40^{\circ }$
$\left(\frac{2}{3} \right) (tan3A)=3$
$\therefore tan3A=\frac{9}{2} $
$\Rightarrow sin3A=\frac{9}{\sqrt{85} } ..........(5^{\circ} <A<30^{\circ })$