Question

спростити вираз ( дуже потрібно) ​

Answer 1

Ответ:

(tg(a)+tg(B))/(ctg(a)+ctg(B))=tg(a)*tg(B)

Пошаговое объяснение:

tg(a)+tg(B)=sin(a)/cos(a)+sin(B)/cos(B)

sin(a)*cos(B)/(cos(a)*cos(B))+sin(B)*cos(a)/(cos(a)*cos(B))

(sin(a)*cos(B)+sin(B)*cos(a))/(cos(a)*cos(B))

tg(a)+tg(B)=sin(a+B)/(cos(a)*cos(B))

ctg(a)+ctg(B)=cos(a)/sin(a)+cos(B)/sin(B)=

cos(a)*sin(B)/(sin(a)*sin(B))+cos(B)*sin(a)/(sin(a)*sin(B)

(cos(a)*sin(B)+cos(B)*sin(a))/(sin(a)*sin(B))

ctg(a)+ctg(B)=sin(a+B)/(sin(a)*sin(B))

(tg(a)+tg(B))/(ctg(a)+ctg(B)) (sin(a+B)/(cos(a)*cos(B)))÷(sin(a+B)/(sin(a)*sin(B)))

(sin(a)*sin(B))/(cos(a)*cos(B))=tg(a)*tg(B)

Answer 2

Ответ:tg∝*tgβ

Пошаговое объяснение:

(tg∝+tgβ)/(ctg∝+ctgβ)=(tg∝+tgβ)/(1/tg∝+1/tgβ)=(tg∝+tgβ)/((tg∝+tgβ)/(tg∝*tgβ))=tg∝*tgβ