Question

Оба номера, если получится)))

Answer 1

Ответ:

Решение

Объяснение:

Answer 2

$1)\frac{tg^{2}\alpha-Ctg^{2}\alpha}{Cos2\alpha }=\frac{\frac{Sin^{2}\alpha }{Cos^{2}\alpha}-\frac{Cos^{2}\alpha}{Sin^{2}\alpha}}{Cos2\alpha}=\frac{\frac{Sin^{4}\alpha-Cos^{4}\alpha}{Sin^{2}\alpha Cos^{2}\alpha}}{Cos2\alpha }=\frac{(Sin^{2}\alpha-Cos^{2}\alpha)( Sin^{2}\alpha + Cos^{2}\alpha)}{Sin^{2}\alpha Cos^{2}\alpha(Cos^{2}\alpha-Sin^{2}\alpha)}=\\\\=-\frac{1}{Sin^{2}\alpha Cos^{2}\alpha} =-\frac{1*4}{4Sin^{2}\alpha Cos^{2}\alpha}=\boxed{-\frac{4}{Sin^{2}2\alpha} }$

$2)\frac{Cos^{6}\alpha-Sin^{6}\alpha}{Cos2\alpha }+\frac{Sin^{2}2\alpha}{4}=\frac{(Cos^{2}\alpha-Sin^{2}\alpha)(Cos^{4}\alpha+Sin^{2}\alpha Cos^{2}\alpha+Sin^{4}\alpha)}{Cos^{2}\alpha-Sin^{2}\alpha}+\frac{Sin^{2}2\alpha }{4}=\\\\=(Cos^{4}\alpha+Sin^{2}\alpha Cos^{2}\alpha+Sin^{4}\alpha)+\frac{Sin^{2}2\alpha }{4}=\\\\=(Cos^{2}\alpha+Sin^{2}\alpha)^{2}-Sin^{2}\alpha Cos^{2}\alpha+\frac{1}{4}*Sin^{2}2\alpha=1-\frac{1}{4}*4Sin^{2}\alpha Cos^{2}\alpha=$

$=1-\frac{1}{4}Sin^{2} 2\alpha+\frac{1}{4}Sin^{2} 2\alpha=\boxed1$