Matemática, perguntado por Dragonboechat, 10 meses atrás

simplifique a expressão tg (pi - x)- tg (pi + x)/ tg (2pi - x)

Soluções para a tarefa

Respondido por Usuário anônimo
6

Explicação passo-a-passo:

\sf tg~(a-b)=\dfrac{tg~a-tg~b}{1+tg~a\cdot tg~b}

\sf tg~(a+b)=\dfrac{tg~a+tg~b}{1-tg~a\cdot tg~b}

Temos que:

\sf tg~(\pi-x)=\dfrac{tg~\pi-tg~x}{1+tg~\pi\cdot tg~x}

\sf tg~(\pi-x)=\dfrac{0-tg~x}{1+0\cdot tg~x}

\sf tg~(\pi-x)=\dfrac{-tg~x}{1}

\sf tg~(\pi-x)=-tg~x

\sf tg~(\pi+x)=\dfrac{tg~\pi+tg~x}{1-tg~\pi\cdot tg~x}

\sf tg~(\pi+x)=\dfrac{0+tg~x}{1-0\cdot tg~x}

\sf tg~(\pi+x)=\dfrac{tg~x}{1}

\sf tg~(\pi+x)=tg~x

\sf tg~(2\pi-x)=\dfrac{tg~2\pi-tg~x}{1+tg~2\pi\cdot tg~x}

\sf tg~(2\pi-x)=\dfrac{0-tg~x}{1+0\cdot tg~x}

\sf tg~(2\pi-x)=\dfrac{-tg~x}{1}

\sf tg~(2\pi-x)=-tg~x

Assim:

\sf \dfrac{tg~(\pi-x)-tg~(\pi+x)}{tg~(2\pi-x)}=\dfrac{-tg~x-tg~x}{-tg~x}

\sf \dfrac{tg~(\pi-x)-tg~(\pi+x)}{tg~(2\pi-x)}=\dfrac{2\cdot (-tg~x)}{-tg~x}

\sf \dfrac{tg~(\pi-x)-tg~(\pi+x)}{tg~(2\pi-x)}=2


Dragonboechat: THX
Respondido por Makaveli1996
0

 \frac{ \tan(\pi - x)  -  \tan(\pi + x) }{ \tan(2\pi - x) }  \\  \frac{ -  \tan(x) -  \tan(x)  }{ -  \tan(x) }  \\  \frac{ - 2 \tan(x) }{ -  \tan(x) }  \\  - 2 \: . \: ( - 1) \\ 2 \: . \: 1 \\ \boxed{\boxed{\boxed{2}}} \\

atte. yrz

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