!!!!!!РЕШИТЕ ПОЖАЛУЙСТА!!!!!!!

Решите задачу:

$\frac{2sin^2(a)}{1+cos(\pi-2a)}-sin^2(a)= \frac{2sin^2(a)}{1-cos(2a)}-sin^2(a)=\\\\ =\frac{2sin^2(a)}{2*\frac{1-cos(2a)}{2}}-sin^2(a)= \frac{2sin^2(a)}{2*sin^2(a)}-sin^2(a)=\\\\ =1-sin^2(a)=cos^2(a)$

$\frac{sin(0.5\pi+2a)}{1-tg^2(a)}-cos^2(a)= \frac{sin(\frac{\pi}{2}+2a)}{1-\frac{sin^2(a)}{cos^2(a)}}-cos^2(a)=\\\\ =\frac{cos(2a)}{\frac{cos^2(a)-sin^2(a)}{cos^2(a)}}-cos^2(a)= \frac{cos(2a)}{\frac{cos(2a)}{cos^2(a)}}-cos^2(a)=\\\\ =cos(2a):\frac{cos(2a)}{cos^2(a)}-cos^2(a)=cos(2a)*\frac{cos^2(a)}{cos(2a)}-cos^2(a)=\\\\ =cos^2(a)-cos^2(a)=0$