Question

A expressão E= tg²(π-x)- tg(π+x)/tg(2π-x), com tg x ≠ 0, é equivalente a: 

Obrigada :)

Answer 1

tg(a-b) = tga - tgb / 1+tga.tgb 
tg(a+b) = tga+tgb/1-tgatgb

tgπ = 0
tg2π = 0

( tgπ - tgx/ 1 + tgπ.tgx )² - tgπ+tgx/ 1-tgπ.tgx  / tg(2π-x)
tg²π -2tgπ.tgx + tg²x / 1+2tgπ.tgx+tg²π.tg²x - tgx / tg(2π-x)
tg²x -tgx / tg2π - tgx / 1+tg2π.tgx
 
tg²x - tgx / -tgx
R: -tgx + 1