sin^2a(1+ctga)+cos^2a(1+tga)-sina=? ctg^2a-cos^2a-ctg^2a*cos^2a=?. Created by lyazkasari. algebra-ru

sin^2a(1+ctga)+cos^2a(1+tga)-sina=? ctg^2a-cos^2a-ctg^2a*cos^2a=?...

July 2022 1 3 Report

sin^2a(1+ctga)+cos^2a(1+tga)-sina=?
ctg^2a-cos^2a-ctg^2a*cos^2a=?

Answers & Comments

georgefedoseev $1) sin^2a(1+ctga)+cos^2a(1+tga)-sina=sin^2a+ctgasin^2a+\\+cos^2a+cos^2atga-sina= sin^2a+cos^2a+ \frac{cosasin^2a}{sina}+ \frac{cos^2asina}{cosa}-\\-sina= 1+2sinacosa-sina=1+sin2a-sina$

$2)ctg^2a-cos^2a-ctg^2acos^2a= \frac{cos^2a}{sin^2a} -cos^2a- \frac{cos^2acos^2a}{sin^2a} =\\= \frac{cos^2a-sin^2acos^2a-cos^2acos^2a}{sin^2a} = \frac{ \frac{1+cos2a}{2} - \frac{(1-cos2a)(1+cos2a)}{4} - \frac{(1+cos2a)^2}{4} }{ \frac{1-cos2a}{2} } =\\= \frac{ \frac{2+2cos2a-1+cos^22a-(1+2cosa+cos^22a)}{4} }{ \frac{1-cos2a}{2} }=\\=\frac{(2+2cos2a-1+cos^22a-1-2cos2a-cos^22a)*2}{4*(1-cos2a)} = \frac{0}{2(1-cos2a)} = 0$

Насчет 2 не знаю, правильно, не правильно, ну вот такой ответ

8 votes Thanks 25