Sin9pi/5*tg5pi/3 решить ,сравнить с нулем выражение

Решите задачу:

$A=sin\frac{9\pi}{5}\cdot tg\frac{5\pi}{3}=sin(\pi +\frac{4\pi}{5})\cdot tg(\pi +\frac{2\pi}{3})=-sin\frac{4\pi}{5}\cdot (-tg\frac{2\pi}{3})=\\\\=sin(\pi -\frac{\pi}{5})\cdot tg(\pi -\frac{\pi}{3})=sin\frac{\pi}{5}\cdot (-tg\frac{\pi}{3})=\\\\=-\sqrt3\cdot sin\frac{\pi}{5}\ \textless \ 0,\; \; t.k.\; \; sin\frac{\pi}{5}\ \textgreater \ 0\; (\; 0\ \textless \ \frac{\pi}{5}\ \textless \ \frac{\pi}{2}\; )\\\\sin\frac{\pi}{5}= \frac{\sqrt{5-\sqrt5}}{2\sqrt2}\; \; \to \; \; A=-\sqrt3\cdot \frac{\sqrt{5-\sqrt5}}{2\sqrt2} =-\frac{\sqrt{3(5-\sqrt5)}}{2\sqrt2}\ \textless \ 0$