Question

1: QUAL DAS RELAÇÕES ABAIXO REPRESENTA O MESMO VALOR DE SEN 330°:
A)SEN 30°
B)-SEN30°
C)COS30°
D)COS60°
2:SE A tg x = a, o valor de tg 160° + tg 340° é?
A)2a
B)-2a
C)a
D)-a
Obs: Preciso da resolução das 2questoes da foto, pfvv.

Answer 1

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$\boxed{\begin{array}{l}\tt 1)\\\sf sen(330^\circ)=-sen(360^\circ-330^\circ)=-sen(30^\circ)\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~B}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\tt 2)~\sf tg(20^\circ)=a\\\sf tg(160^\circ)+tg(340^\circ)\\\sf tg(160^\circ)=-tg(180^\circ-160^\circ)=-tg(20^\circ)=-a\\\sf tg(340^\circ)=-tg(360^\circ-340^\circ)=-tg20^\circ=-a\\\sf tg(160^\circ)+tg(340^\circ)=-a-a=-2a\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~B}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\tt 3)~\sf sen(x)=-\dfrac{3}{5},x\in 3^o~quadrante.\\\sf cosec(x)=-\dfrac{5}{3}\implies cosec^2(x)=\dfrac{25}{9}\\\sf cotg^2(x)=cosec^2(x)-1\\\sf cotg^2(x)=\dfrac{25}{9}-1\\\sf cotg^2(x)=\dfrac{16}{9}\\\sf cotg(x)=\sqrt{\dfrac{16}{9}}\\\sf cotg(x)=\dfrac{4}{3}\\\sf tg(x)=\dfrac{3}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\tt 4)~\sf \dfrac{7\pi}{2}=\dfrac{4\pi}{2}+\dfrac{3\pi}{2}=2\pi+\dfrac{3\pi}{2}\\\sf \dfrac{25\pi}{6}=\dfrac{24\pi}{6}+\dfrac{\pi}{6}=2\cdot2\pi+\dfrac{\pi}{6}\\\sf sen\bigg(\dfrac{7\pi}{2}\bigg)=sen\bigg(\dfrac{3\pi}{2}\bigg)=-1\\\sf sen\bigg(\dfrac{25\pi}{6}\bigg)=sen\bigg(\dfrac{\pi}{6}\bigg)=\dfrac{1}{2}\\\sf sen\bigg(\dfrac{7\pi}{2}\bigg)-sen\bigg(\dfrac{25\pi}{6}\bigg)=-1-\dfrac{1}{2}=-\dfrac{3}{2}\end{array}}$