Question

Помогите решить алгебру:))))))

Answer 1

$sin^2a+sin^2b-sin^2a*sin^2b+cos^2a*cos^2b =\ sin^2a+sin^2b-sin^2a*sin^2b+(1-sin^2a)(1-sin^2b)=\ sin^2a+sin^2b-sin^2a*sin^2b+(1-sin^2b-sin^2a+sin^2a*sin^2b)=\ -sin^2a*sin^2b+1+sin^2a*sin^2b=1$



$frac{2sinx-cosx}{2cosx+sinx}=3\ \ 2sinx-cosx=6cosx+3sinx\ -7cosx=sinx\ \ tgx=frac{sinx}{cosx}=frac{-7cosx}{cosx}=-7$


$frac{sin^4a+2cosa*sina-cos^4a}{2cos^2a-1}=frac{sin2a-cos2a}{cos2a}=tg2a-1$