Question

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

Answer 1

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:

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

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

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

Answer 2

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

atte. yrz