Question

Тригонометрия. cpoчно! 1) (sin^2x - cos^2x)^2 + sin^2*2x /1-tgx) - (1/1+tgx))*(cos^2*x) 3) (tg^3*x-tg^3*y)/((1+tgxtgy)(tg^2*x+tgxtgy+tg^2*y))

Answer 1

$( sin^{2} x - cos^{2} x)^{2} + sin^{2} 2 x= cos^{2} 2x + sin^{2} 2x= 1$
$\frac{1}{1-tgx} - \frac{1}{1+tgx} * cos2x = \frac{1+tgx-1+tgx}{(1-tgx)(1+tgx)} *cos2x = \frac{2 tgx}{1-tg^{2}x} * cos2x = tg2x* cos2x = \frac{sin2x * cos2x}{cos2x} = sin2x$
$\frac{tg^{3} x-tg^{3}y}{(1+tgx tgy)*(tg^{2}+tgx tgy+tg^{2})} = \frac{(tgx-tgy)*(tg^{2}+tgx tgy+tg^{2})}{(1+tgx tgy)*(tg^{2}+tgx tgy+tg^{2})} = \frac{(tg x-tg y)}{(1+tg x tg y)} = tg (x-y)$