Question

A forma simplificada da expressão abaixo é:
(sec x- cos x) (cossec x - sen x) (tg x + cotg x)
(A) 1.
(B) tg x.
(C) cos x.
(D) sen x.
(E) –1.

Answer 1

Explicação passo-a-passo:

$\sf (sec~x-cos~x)\cdot(cossec~x-sen~x)\cdot(tg~x+cotg~x)$

$\sf =\Big(\dfrac{1}{cos}-cos~x\Big)\cdot\Big(\dfrac{1}{sen~x}-sen~x\Big)\cdot\Big(\dfrac{sen~x}{cos~x}+\dfrac{cos~x}{sen~x}\Big)$

$\sf =\Big(\dfrac{1-cos^2~x}{cos~x}\Big)\cdot\Big(\dfrac{1-sen^2~x}{sen~x}\Big)\cdot\Big(\dfrac{sen^2~x+cos^2~x}{sen~x\cdot cos~x}\Big)$

Pela relação fundamental da trigonometria:

=> $\sf sen^2~x+cos^2~x=1$

=> $\sf sen^2~x=1-cos^2~x$

=> $\sf cos^2~x=1-sen^2~x$

Assim:

$\sf \Big(\dfrac{1-cos^2~x}{cos~x}\Big)\cdot\Big(\dfrac{1-sen^2~x}{sen~x}\Big)\cdot\Big(\dfrac{sen^2~x+cos^2~x}{sen~x\cdot cos~x}\Big)$

$\sf =\dfrac{sen^2~x}{cos~x}\cdot\dfrac{cos^2~x}{sen~x}\cdot\dfrac{1}{sen~x\cdot cos~x}$

$\sf =\dfrac{sen^2~x\cdot cos^2~x}{sen^2~x\cdot cos^2~x}$

$\sf =1$

Letra A