.... Задание прикреплено....

Решите задачу:

$2)\frac{1+tg^{2}(-\frac{\pi }{6})}{1+Ctg^{2}(-\frac{\pi }{6})}=\frac{1+tg^{2}\frac{\pi }{6}}{1+Ctg^{2}\frac{\pi }{6}}= \frac{\frac{1}{Cos^{2}\frac{\pi }{6}}}{\frac{1}{Sin^{2}\frac{\pi }{6}} }=\frac{Sin^{2}\frac{\pi }{6}}{Cos^{2}\frac{\pi }{6}}=tg^{2}\frac{\pi }{6}=(\frac{1}{\sqrt{3}} )^{2}=\frac{1}{3}$

$4)\frac{3-Sin^{2}(-\frac{\pi }{3})-Cos^{2}(-\frac{\pi }{3})}{2Cos(-\frac{\pi }{4}) }=\frac{3-(Sin^{2}\frac{\pi }{3}+Cos^{2}\frac{\pi }{3})}{2Cos\frac{\pi }{4} }=\frac{3-1}{2*\frac{\sqrt{2} }{2} } =\frac{2}{\sqrt{2} }=\sqrt{2}$

$2)Cos\alpha-Ctg\alpha*Sin(-\alpha)=Cos\alpha+\frac{Cos\alpha }{Sin\alpha}*Sin\alpha =Cos\alpha +Cos\alpha=2Cos\alpha\\\\4)tg(-\alpha )*Ctg(-\alpha)+Cos^{2}(-\alpha)+Sin^{2}\alpha =tg\alpha*Ctg\alpha+(Cos^{2}\alpha+Sin^{2}\alpha)=1+1=2$