Let $\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$
My attempt :
\begin{align*}
\dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\
\tan2A=2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)
\end{align*}
and
\begin{align*}
\tan6A&=\tan(2A-60^\circ)\tan2A\tan(2A+60^\circ)\\
&=(\dfrac{\tan2A-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(\tan2A)(\dfrac{\tan2A+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})
\end{align*}
give
$$\tan6A=(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ))(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})$$
This method is incredibly long, there may be better way to deal with this problem.
Best Answer
Note that: \begin{array}{} \dfrac{\tan2A}{2}&=&\sin^2160^\circ-\sin160^\circ\sin220^\circ+\sin^2220^\circ\\ \dfrac{\tan2A}{2}&=&(\sin160^\circ-\sin220^\circ)^2+\sin160^\circ\sin220^\circ\\ \dfrac{\tan2A}{2}&=&(\sin20^\circ+\sin40^\circ)^2-\sin20^\circ\sin40^\circ\end{array} Apply product-sum and sum-product \begin{array}{}\dfrac{\tan2A}{2}&=&(2\sin30^\circ\cos10^\circ)^2-\dfrac{\cos20^\circ-\cos60^\circ}{2}\\ \dfrac{\tan2A}{2}&=&\cos^210^\circ-\dfrac{1-2\sin^210^\circ}{2}+\dfrac{1}{4}\\ \dfrac{\tan2A}{2}&=&\cos^210^\circ+\sin^210^\circ-\dfrac{1}{4}\\ \dfrac{\tan2A}{2}&=&\dfrac{3}{4}\\ \tan(2A)&=&\dfrac{3}{2} \end{array}