Докажите тождество (1-2sin^2a/cosa+sina)/(1+2cos^2a/sina-cosa)= 2cos a

[latex] \frac{1-2sin^2a}{cosa+sina} : \frac{1-2cos^2a}{sina-cosa}=\\\\\star \; \; 2sin^2a=1-cos2a\; \; \; \Rightarrow \quad 1-2sin^2a=cos2a \\\\\star \; \; 2cos^2a=1+cos2a\; \; \; \Rightarrow \quad 1-2cos^2a=-cos2a\\\\=\frac{cos2a}{sina+cosa} \cdot \frac{sina-cosa}{-cos2a} = -\frac{sina-cosa}{sina+cosa} =[ \frac{:cosa}{:cosa} ]=\\\\=- \frac{tga-1}{tga+1}= \frac{1-tga}{1+tga}=[\; 1=tg\frac{\pi}{4}]= \frac{tg\frac{\pi}{4}-tga}{1+tg\frac{\pi}{4}\cdot tga} =tg(\frac{\pi}{4}-a)[/latex]