Question

(1-sina*tga*cos)(1+tg(квадрат)a)

Answer 1

$(1-sinalphacdot tgalphacdotcosalpha)(1+tg^2alpha)=(1-sinalphacdotfrac{sinalpha}{cosalpha}cdotcosalpha)(1+tg^2alpha)=\=(1-sin^2alpha)(1+tg^2alpha)=cos^2alpha(1+frac{sin^2alpha}{cos^2alpha})=cos^2alpha+sin^2alpha=1$