IME / ITA ⇒ (Simulado-IIta) Trigonometria

[botelho]

Se Tang(x+y+z)=0, reduza [tex3]\frac{tangx+tangy+tangz}{tangx.tangz}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{tangx+tangy+tangz}{tangx.tangz}[/tex3]

.
a)coty
b)tang(x+y)
c)tangy
d)tangy.cotgy
e)cotg(x+y)

Resposta

---

b

[rcompany]

[tex3]
\begin{align}
\dfrac{\tan x+ \tan y +\tan z}{\tan x+\tan z}&=\dfrac{\tan x+\tan y -\tan(x+y)}{-\tan x \tan (x+y)}\quad\text{já que }\tan(x+y+z)=0\implies\tan(x+y)=-\tan z\\
&=\dfrac{(1-\tan x\tan y)\tan(x+y)-\tan (x+y)}{-\tan x\tan(x+y)}\quad\text{já que }\tan x+\tan y=\tan(x+y)(1-\tan x\tan y)\\
&=\dfrac{-\tan x\tan y\tan(x+y)}{-\tan x\tan(x+y)}\\
&=\tan y
\end{align}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\begin{align}
\dfrac{\tan x+ \tan y +\tan z}{\tan x+\tan z}&=\dfrac{\tan x+\tan y -\tan(x+y)}{-\tan x \tan (x+y)}\quad\text{já que }\tan(x+y+z)=0\implies\tan(x+y)=-\tan z\\
&=\dfrac{(1-\tan x\tan y)\tan(x+y)-\tan (x+y)}{-\tan x\tan(x+y)}\quad\text{já que }\tan x+\tan y=\tan(x+y)(1-\tan x\tan y)\\
&=\dfrac{-\tan x\tan y\tan(x+y)}{-\tan x\tan(x+y)}\\
&=\tan y
\end{align}
[/tex3]

Gabarito aparentemente errado.