Sin2x+cos2x=0 решите подробно

sin2x = 2tgx / (1 + tg2 x) ;

cos2x = (1 - tg2x) / (1 + tg2 x) ;

Sin2x+cos2x = 2tgx / (1 + tg2x) + (1 - tg2x) / (1 + tg2x) = (1 - tg2x+2tgx) / (1 + tg2x) ;

tg2x=2tgx / (1-tg^2 x) ;

1 + tg2x = (1-tg^2 x) / (1-tg^2 x) + 2tgx / (1-tg^2 x) = - (tgx-1) ^2 / (tgx-1) (tgx+1) = (1-tgx) / (1+tgx) ;

1 - tg2x+2tgx=1+2tgx-2tgx / (1-tgx) (tgx+1) = ((1-tgx) (tgx+1) + 2tgx (1-tgx) (tgx+1) - 2tgx) / (1-tgx) (tgx+1) ;

((1-tgx) (tgx+1) + 2tgx (1-tgx) (tgx+1) - 2tgx) / (1-tgx) ^2;

(tg^2 x - 2tgx + 1) / (1-tgx) ^2;

(tgx-1) ^2 / (1-tgx) ^2=1 тк (1-tgx) ^2 = (tgx-1) ^2