Срочно помогите упр 710(1,3) !!!!!!

Решите задачу:

$1)\;\frac{\cos(-\alpha)+3\cos180^o+\cos(180^o-\alpha)}{\tan(-60^o)-\cos(-120^o)-\tan(-420^o)}=\frac{\cos\alpha+3\cdot(-1)-\cos\alpha}{-\tan60^o-\cos120^o+\tan420^o}=\\=\frac{-3}{-\tan60^o+\frac12+\tan(360^o+60^o)}=\frac{-3}{-\tan60^o+\frac12+\tan60^o}=\frac{-3}{\frac12}=-6$

$3)\;\cos(-\alpha)\tan(\alpha-180^o)-\tan(-\alpha):\cos(180^o-\alpha)=\\=\cos\alpha\cdot(-\tan(180^o-\alpha))+\tan\alpha:(-\cos\alpha)=\\=\cos\alpha\tan\alpha-\tan\alpha:\cos\alpha=\cos\alpha\cdot\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\alpha}{\cos^2\alpha}=\\=\sin\alpha\cdot\left(1-\frac1{\cos^2\alpha}\right)=-\sin\alpha\left(\frac1{\cos^2\alpha}-1\right)=-\sin\alpha\cdot\frac{1-\cos^2\alpha}{\cos^2\alpha}=\\=-\sin\alpha\cdot\frac{\sin^2\alpha}{\cos^2\alpha}=-\sin\alpha\tan^2\alpha$