Question

sa se verifice egalitatile


$arctg \frac{1}{2} + arctg \frac{1}{3} = \frac{ \pi }{4}$

Answer 1

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