Нужна помощь. (sin^2B - cos^2B + 1)/sin^2B cos80 + cos40 - cos20= 0 (кроме нуля все в градусах) Буду благодарен, если поможете.

[latex] \frac{ sin^{2} \beta -cos^{2} \beta +1 }{sin ^{2} \beta } = \frac{ sin^{2} \beta -(1- sin^{2} \beta) +1 }{sin^{2} \beta } = \frac{ sin^{2} \beta -1+ sin^{2} \beta +1 }{ sin^{2} \beta } = \frac{2* sin^{2} \beta }{ sin^{2} \beta } =2[/latex] cos80°+cos40°-cos20°=(cos80°+cos40°)-cos20°= [latex]=2*cos \frac{ 80^{0}+ 40^{0} }{2}*cos \frac{ 80^{0} - 40^{0} }{2} -cos20°=[/latex] =2*cos60°*cos20°-cos20°=2*cos20° *(2*cos60°-1)=2*cos20° *(1/2-1)=2*(-1/2)*cos20°=-cos20°