Question

До утра нужно сделать оба задания, надеюсь кто-то поможет. Спасибо заранее.

Answer 1

$1a)frac{Cos80^{o} }{Sin40^{o}-Cos40^{o}}=frac{Cos^{2}40^{o} -Sin^{2}40^{o}}{Sin40^{o}-Cos40^{o}}=frac{(Cos40^{o}-Sin40^{o})(Cos40^{o}+Sin40^{o})}{Sin40^{o}-Cos40^{o}}=-(Cos40^{o}+Sin40^{o})$

$frac{Sin2alpha+Sin8alpha}{Cos2alpha-Cos8alpha}=frac{2Sinfrac{2alpha+8alpha}{2}Cosfrac{2alpha-8alpha}{2}}{-2Sinfrac{2alpha+8alpha}{2}Sinfrac{2alpha-8alpha}2}} =frac{Sin5alpha Cos3alpha}{Sin5alpha Sin3alpha}=frac{Cos3alpha }{Sin3alpha }=Ctg3alpha$

$2a)frac{1}{1-tgalpha }-frac{1}{1+tgalpha }=frac{1+tgalpha-1+tgalpha}{(1-tgalpha)(1+tgalpha)}=frac{2tgalpha }{1-tg^{2}alpha}=tg2alpha\\tg2alpha=tg2alpha$

тождество доказано

2б)

$(frac{Cosalpha }{Cos4alpha } -frac{Sinalpha }{Sin4alpha })*frac{Cos6alpha-Cos10alpha}{Sin3alpha }=frac{Cosalpha Sin4alpha-Cos4alpha Sinalpha}{Sin4alpha Cos4alpha}*frac{-2Sinfrac{6alpha+10alpha}{2}Sinfrac{6alpha-10alpha}{2}}{Sin3alpha }=frac{Sin(4alpha-alpha)}{Sin4alpha Cos4alpha}*frac{2Sin8alpha Sin2alpha}{Sin3alpha }=frac{2Sin3alpha }{Sin8alpha }*frac{2Sin8alpha Sin2alpha }{Sin3alpha }=4Sin2alpha\\4Sin2alpha=4Sin2alpha$

Тождество доказано