Question

(Fgv 2008) Sendo p = 1/2 e (p + 1) ⋅ (q + 1) = 2, então a medida de arctg p + arctg q , em radianos, é

Answer 1

Boa noite

Se p= 1/2 e (p+1)+(q+1)=2 temos (1/2+1)+(q+1)=2 ⇒q= 1/3

arc tg p = x ⇒   p=tg x = 1/2      e   arc tg q=y ⇒  q= tg y = 1 / 3

 $tg(x+y)= \dfrac{tgx+tgy}{1-tgx*tgy}\Rightarrow tg(x+y)= \dfrac{ \frac{1}{2}+ \frac{1}{3} }{1- \frac{1}{2}* \frac{1}{3} } = \dfrac{ \frac{5}{6} }{ \frac{5}{6} } \\ \\ tg(x+y)=1\Rightarrow x+y= \pi /4 \\ \\ arctgp+arctgq = \dfrac{ \pi }{4}$