Упрастите...........

1. $\frac{4cos2a}{sin^22a(ctg^2a-tg^2a)} = \frac{4cos2a}{(sin2a)^2( \frac{cos^2a}{sin^2a} - \frac{sin^2a}{cos^2a}) } = \frac{4cos2a}{4sin^2acos^2a*( \frac{cos^4a-sin^4a}{sin^2acos^2a}) }$ $= \frac{4cos2a}{4*(cos^4a-sin^4a)} = \frac{cos2a}{(cos^2a-sin^2a)(cos^2a+sin^2a)} = \frac{(cos^2a-sin^2a)}{(cos^2a-sin^2a)*1} = 1$

2. $\frac{cos^4a-sin^4a-cos^2a}{2(cosx-1)} - cos^2 \frac{a}{2} = \frac{(cos^4a-sin^4a)-cos^2a}{-2(1-cosa)} - cos^2 \frac{a}{2}=$ $= \frac{(cos^2a-sin^2a)(cos^2a+sin^2a)-cos^2a}{-2(1-cosa) } - cos^2 \frac{a}{2} = \frac{cos^2a-sin^2a-cos^2a}{-2(1-cosa)} - cos^2 \frac{a}{2} =$ $= \frac{-sin^2a}{-2(1-cosa)} - cos^2 \frac{a}{2} = \frac{sin^2a}{2(1-cosa)} - \frac{1+cosa}{2}$ $= \frac{sin^2a}{2(1-cosa)}- \frac{(1+cosa)(1-cosa)}{2(1-cosa)} = \frac{sin^2-(1^2-cos^2a)}{2(1-cosa)} = \frac{sin^2a+cos^a-1}{2(1-cosa)} =$ $\frac{1-1}{2(1-cosa)} = 0$