Sejam \(x, y \in (o\frac{\pi}{2})\), tais que \(cos(x) = \frac{4} {5}\) e \(sen(y) = { \large 5 \over 13}\). Podemos concluir que \(tg(x + y)\) é igual a:

Gabarito

Errado

Comentário do professor

Como \(x,y\in(0,\tfrac{\pi}{2})\), as razões trigonométricas são positivas.

Seno de x: \( \sin x=\sqrt{1-\cos^2x}=\sqrt{1-(4/5)^2}=\sqrt{1-16/25}=\sqrt{9/25}=3/5\).
Cosseno de y: \( \cos y=\sqrt{1-\sin^2y}=\sqrt{1-(5/13)^2}=\sqrt{1-25/169}=\sqrt{144/169}=12/13\).

Tangentes: \( \tan x=\dfrac{\sin x}{\cos x}=\dfrac{3/5}{4/5}=\tfrac{3}{4}\), \( \tan y=\dfrac{\sin y}{\cos y}=\dfrac{5/13}{12/13}=\tfrac{5}{12}\).

Tangente da soma: \( \tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\tfrac{3}{4}+\tfrac{5}{12}}{1-\tfrac{3}{4}\cdot\tfrac{5}{12}}=\dfrac{\tfrac{14}{12}}{1-\tfrac{5}{16}}=\dfrac{14/12}{11/16}=\dfrac{14}{12}\cdot\dfrac{16}{11}=\dfrac{56}{33}.\)