Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$ 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.
 A: 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}
A: Let me post a different route to the simplication, as OP has begun.
So we have
$$
\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ=
\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos^2 20^\circ)
$$
I claimed for all $\theta$,
$$
\sin^2\theta(1+2\cos\theta+4\cos^2\theta)=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta-\cos4\theta)
$$
Proof of claim:
\begin{align*}
LHS&=\sin^2\theta(1+2\cos\theta+4\cos^2\theta)\\
&=\frac12(1-\cos 2\theta)(3+2\cos\theta+2\cos2\theta)\\
&=\frac12(3+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- 2\cos^22\theta)\\
&=\frac12(2+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- \cos4\theta)\\
&=\frac12(2+\cos\theta-\cos2\theta-(4\cos^3\theta-3\cos\theta)- \cos4\theta)\\
&=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta- \cos4\theta)\\
&=RHS\quad\checkmark
\end{align*}
So with $\theta=20^\circ$, this is
\begin{align*}
&\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos 40^\circ)\\
&=\frac12(2+\cos 20^\circ-\cos40^\circ-\cos60^\circ- \cos80^\circ)\\
&=\frac12\left(\frac32+\cos 20^\circ-\cos40^\circ-\cos80^\circ\right)\\
&=\frac12\left(\frac32+\cos 20^\circ-2\cos60^\circ\cos20^\circ\right)\\
&=\frac34.
\end{align*}
So $\tan 2A=\frac32$ and
$$
\tan 6A=\tan 3(2A)=\frac{3\tan 2A-\tan^3 2A}{1-3\tan^2 2A}=\dots
$$
