Помогите пожалуйста решить 1 вариант

α - угол третьей четверти поэтому Sinα < 0

$Sin\alpha=- \sqrt{1-Cos^{2}\alpha}=-\sqrt{1-(-0,6)^{2}}=-\sqrt{1-0,36}=-\sqrt{0,64}=-0,8\\tg(\frac{\pi }{2}+\alpha)=-Ctg\alpha=-\frac{Cos\alpha }{Sin\alpha }=-\frac{-0,6}{-0,8}=-0,75\\2)\frac{2Sin^{2}\alpha*Ctg\alpha}{Cos^{2}\alpha-Sin^{2}\alpha}=\frac{2Sin^{2}\alpha*\frac{Cos\alpha }{Sin\alpha }}{Cos2\alpha} =\frac{2Sin\alpha Cos\alpha}{Cos2\alpha }=\frac{Sin2\alpha }{Cos2\alpha}=tg2\alpha\\\\tg2\alpha=tg2\alpha \\\frac{2Sin^{2} \alpha }{tg2\alpha*tg\alpha}=Cos^{2}\alpha-Sin^{2}\alpha$

$\frac{2Sin^{2}\alpha}{tg2\alpha*\frac{Sin\alpha}{Cos\alpha }}=Cos2\alpha\\\frac{2Sin\alpha Cos\alpha}{tg2\alpha }=Cos2\alpha\\\frac{Sin2\alpha }{\frac{Sin2\alpha }{Cos2\alpha} }=Cos2\alpha \\Cos2\alpha=Cos2\alpha$

$3)Sin240^{0} =Sin(180^{0}+60^{0} )=-Sin60^{0}=-\frac{\sqrt{3}}{2} \\Cos\frac{3\pi }{4}=Cos(\pi-\frac{\pi }{4})=-Cos\frac{\pi }{4}=-\frac{\sqrt{2}}{2}\\ Ctg(-\frac{\pi }{6})=-Ctg\frac{\pi }{6} =-\sqrt{3}$