Question

Sin60°- 8tg45°- cos30°-8tg135°;

Answer 1

$Sin60^{o}-8tg45^{o}-Cos30^{o}-8tg135^{o}=\frac{\sqrt{3} }{2}-8*1-\frac{\sqrt{3}}{2}-8tg(180^{o}-45^{o})=-8+8tg45^{o}=-8+8*1=-8+8=0\\\\Otvet:\boxed{0}$