Estude! Com Prof. Caju

Sabendo-se que [tex3]\tan x=a[/tex3] e [tex3]\tan y=b[/tex3] ; é possivel reescrever [tex3]Z=\frac{sen2x+sen2y}{sem2x-sen2y}[/tex3] como:

A) [tex3]\left(\frac{1-ab}{1+ab}\right)\cdot\left(\frac{a-b}{a+b}\right)[/tex3]
B) [tex3]\left(\frac{1+ab}{1-ab}\right)\cdot\left(\frac{a-b}{a+b}\right)[/tex3]
C) [tex3]\left(\frac{1-ab}{1+ab}\right)\cdot\left(\frac{a+b}{a-b}\right)[/tex3]
D) [tex3]\left(\frac{1+ab}{1-ab}\right)\cdot\left(\frac{-a+b}{a-b}\right)[/tex3]
E) [tex3]\left(\frac{1+ab}{1-ab}\right)\cdot\left(\frac{a+b}{a-b}\right)[/tex3]

Resposta

GAB: E

JohnnyEN,

[tex3]sen2x+sen2y = 2sen(x+y)cos(x-y)(I)\\
sen2x-sen2y = 2sen(x-y)cos(x+y)(II)\\
(I\div II): \frac{tg(x+y)}{tg(x-y) }=\frac{\frac{tg(x)+tg(y)}{1-tg(x)tg(y)}}{\frac{tg(x)-tg(y)}{1+tg(x)tg(y)}}=\\
\frac{a+b}{1-ab}.\frac{1+ab}{a-b}=\boxed{\frac{a+b}{a-b}.\left(\frac{1+ab}{1-ab}\right)}[/tex3]

Estude! Com Prof. Caju

(Escola naval)-Trigonometria

(Escola naval)-Trigonometria

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