Решите тригонометрическое выражение

3/2

Пошаговое объяснение:

tg(a-b)=(tga-tgb)/(1+tga·tgb)

tg(3π/4)= -1

sin²α+cos²α=1

sin2α=2·sinα·cosα

tg(3π/4-α)=(tg3π/4-tgα)/(1+tg3π/4·tgα)=(-1-tgα)/(1+(-1)·tgα)=

= -(1+tgα)/(1-tgα)=-(cosα+sinα)/(cosα-sinα)

tg²(3π/4-α)=(-(cosα+sinα)/(cosα-sinα))²=

=(cos²α+2·cosα·sinα+sin²α)/(cos²α-2·cosα·sinα+sin²α)=(1+sin2α)/(1-sin2α)

3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)

sin2α=-1/3

3tg²(3π/4-α)=3·(1+sin2α)/(1-sin2α)=3·(1+(-1/3))/(1-(-1/3))=3·(2/3)/(1+1/3)=

=3·(2/3)/(4/3)=3·(2/3)·(3/4)=3/2