aplicam functioa tangenta in stanga si in dreapta, functia tangenta fiind bijectiva si deci inversabila pe (-π/2;π/2)->R
iar 1/2 s1/3 ∈(-π/2; π/2)
tg(arctg1/2+arctg1/3) =tgπ/4
aplicam formyul a tg (α+β)= (tgα+tgβ)/(1-tgαtgβ)
in cazul nostru
[tg(arctg1/2) +tg(arctg1/3)]/[(1-tg(arctg1/2) *tg*arctg1/3)]=1
(1/2+1/3) /(1-1/6)=1
(5/6)/(5/6)=1
1=1 adevarat