Question

Gente, eu preciso que alguém responda a 1 e a 3 desse anexo, pois não estou conseguindo:
Obs: a 2 não é necessária.

1) A expressão:

y = $\frac{(secx-tgx)(secx+tgx)}{(1-sen^{2}x)(cotgx-cossecx)(cotgx+cossecx)}$

equivale a:
a) -sec²x
b) cossec²x
c) -cossec²x
d) cos²x
e) -cos²x


2) Sabendo que senx = $\frac{1}{3}$ e

$\frac{\pi }{2} \ \textless \ x\ \textless \ \pi$, o valor de $\frac{cossecx-secx}{cotgx-1}$ é:

a) $\frac{3\sqrt{2}}{4}$
b) $\frac{2\sqrt{2}}{3}$
c) $\frac{-3\sqrt{2}}{4}$
d) $\frac{-2\sqrt{2}}{3}$
e) 3


3) A expressão $\frac{sen(180-a)tg(-a) tg 255}{sen570.cos(180+a) cotg(90-a)}$

a) 0
b) sena
c) -1
d) -2tga
e) cosa

dcc4b4642f2853a9dd1fc4aac4bbec9e.pdf

Answer 1

Resposta:

a) Alternativa a)

c) Alternativa d)

Explicação passo-a-passo:

a)

$\displaystyle y=\frac{(secx-tgx)(secx+tgx)}{(1-sen^2x)(cotgx-cossecx)(cotgx+cossecx)} \\\\\\\displaystyle y=\frac{(sec^2x-tg^2x)}{(cos^2x)(cotg^2x-cossec^2x)} \\\\\\y=\frac{\frac{1}{cos^2x}-\frac{sen^2x}{cos^2x}}{(cos^2x)(\frac{cos^2x}{sen^2x}-\frac{1}{sen^2x} )} =\frac{\frac{1-sen^2x}{cos^2x} }{(cos^2x)(\frac{cos^2x-1}{sen^2x}) } =\frac{\frac{cos^2x}{cos^2x} }{-(cos^2x)(\frac{1-cos^2x}{sen^2x}) } \\\\\\y=\frac{1}{-(cos^2x)(\frac{sen^2x}{sen^2x} )}=-\frac{1}{(cos^2x)} =-sec^2x$

b)

$\displaystyle y=\frac{sen(180^{\circ}-a).tg(-a).tg225^{\circ}}{sen570^{\circ}.cos(180^{\circ}+a)(cotg(90^{\circ}-a))} \\\\\\y=\frac{(sen180^{\circ}.cosa-senacos180^{\circ}).tg(-a).\frac{sen225^{\circ}}{cos255^{\circ}} }{-sen30^{\circ}.(cos180^{\circ}cosa+sen180^{\circ}sena).(\frac{cos(90^{\circ}-a)}{sen(90^{\circ}-a)} )}$

$\displaystyle\\y=\frac{sena.tg(-a).\frac{sen(270^{\circ}-45^{\circ})}{cos(270^{\circ}-45^{\circ})} }{-sen30^{\circ}.(-cos(a)).\frac{cos90^{\circ}cosa+sen90^{\circ}sena}{sen90^{\circ}cosa-senacos90^{\circ}} } \\\\\\y=\frac{sena.tg(-a).\frac{sen270^{\circ}cos45^{\circ}-sen45^{\circ}cos270^{\circ}}{cos270^{\circ}cos45^{\circ}+sen270^{\circ}sen45^{\circ}} }{sen30^{\circ}.cosa.\frac{0+sena}{cosa-0} }$

$\displaystyle\\y=\frac{sena.tg(-a).\frac{-cos45^{\circ}-0}{0-sen45^{\circ}} }{sen30^{\circ}.cosa.sena}}\\\\\\y=\frac{tg(-a)}{sen30^{\circ}} =\frac{tg(-a)}{\frac{1}{2} }=2tg(-a)$

$\displaystyle\\y=2tg(0^{\circ}-a)=2.\frac{tg0^{\circ}-tga}{1+tg0^{\circ}.tga} =2.\frac{(0-tga)}{1+0}=-2tga$