464(1,3) 465(1) заранее спасибо!)))

Решите задачу:

$1)\quad \frac{1-2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}-1}=\frac{1-(1-cosx)}{(1+cosx)-1}=\frac{cosx}{cosx}=1\\\\2)\quad cos^4a-sin^4a-cos2a=\\\\=(\underbrace {cos^2a-sin^2a}_{cos2a})(\underbrace {cos^2a+sin^2a}_{1})-cos2a=cos2a-cos2a=0$

$3)\quad \frac{1+cos42^\circ }{1-cos42^\circ } = \frac{2cos^284^\circ}{2sin^284^\circ } =ctg^284^\circ=tg^2(90^\circ-6^\circ )=tg^26^\circ\\\\\\Formylu:\quad 1-cosa=2sin^2\frac{a}{2}\; \; ;\; \; \; 1+cosa=2cos^2\frac{a}{2}\\\\ili\\\\ sin^2\frac{a}{2}=\frac{1-cosa}{2}\; \; \; ;\; \; \; cos^2\frac{a}{2}= \frac{1+cosa}{2}$