Question

Упрастите...........

Answer 1

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$