Докажите тождество1)(2-sina)(2+sina)+(2-cosa)(2+cosa)=72)sin^4a-cos^4a=sin^2a-cos^2a.

[latex](2-sin\alpha)(2+sin\alpha)+(2-cos\alpha)(2+cos\alpha)=7\\2^2-sin^2\alpha+2^2-cos^2\alpha=7\\sin^2\alpha+cos^2\alpha=1\\\\sin^4\alpha-cos^4\alpha=sin^2\alpha-cos^2\alpha\\sin^4\alpha-sin^2\alpha-cos^4\alpha+cos^2\alpha=0\\sin^2\alpha(sin^2\alpha-1)-cos^2\alpha(cos^2\alpha-1)=0\\sin^2\alpha*(-cos^2\alpha)-cos^2\alpha*(-sin^2\alpha)=0\\-sin^2\alpha cos^2\alpha+sin^2\alpha cos^2\alpha=0[/latex]