Question

1) Calcule o valor da expressão abaixo, sabendo que senx = 3/5: z =( sec x – cos x ) x (tg x + cotg x).
2) Se sen x= √6/3 , com 0 < x < 90º, calcule tgx:

3) Sendo senx=1/2 , com 90° < x < 180°, calcule o valor de: a)cosx= b)secx= c)cossecx= d)tgx= e) cotgx=.

4) Sendo senx=1/3, com 0≤x≤90º, calcule o valor da expressão: y=(senx.cos⁡x)/(sen²x).

Answer 1

$\boxed{\begin{array}{l}\rm 1)~\sf sen(x)=\dfrac{3}{5}\implies sen^2(x)=\dfrac{9}{25}\\\sf cos^2(x)=\dfrac{25}{25}-\dfrac{9}{25}=\dfrac{16}{25}\\\sf cos(x)=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\\sf sec(x)=\dfrac{5}{4}\\\sf tg(x)=\dfrac{\frac{3}{\backslash\!\!\!5}}{\frac{4}{\backslash\!\!\!5}}=\dfrac{3}{4}\\\sf cotg(x)=\dfrac{4}{3}\\\sf z=\dfrac{sec(x)-cos(x)}{tg(x)+cotg(x)}\\\sf z=\dfrac{\frac{5}{4}-\frac{4}{5}}{\frac{3}{4}+\frac{4}{3}}\\\sf z=\dfrac{\frac{25-16}{20}}{\frac{9+16}{12}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf z=\dfrac{\frac{9}{20}}{\frac{25}{12}}\\\\\sf z=\dfrac{\diagdown\!\!\!\!9^3}{\diagdown\!\!\!\!\!\!20_4}\cdot\dfrac{\diagdown\!\!\!\!\!\!25^5}{\diagdown\!\!\!\!\!\!12_4}\\\sf z=\dfrac{15}{16}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2)~\sf sen(x)=\dfrac{\sqrt{6}}{3}\implies sen^2(x)=\dfrac{6}{9}\\\sf cos^2(x)=\dfrac{9}{9}-\dfrac{6}{9}=\dfrac{3}{9}\\\sf cos(x)=\sqrt{\dfrac{3}{9}}=\dfrac{\sqrt{3}}{3}\\\sf tg(x)=\dfrac{sen(x)}{cos(x)}\\\sf tg(x)=\dfrac{\frac{\sqrt{6}}{\diagdown\!\!\!\!3}}{\frac{\sqrt{3}}{\diagdown\!\!\!\!3}}\\\sf tg(x)=\dfrac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 3)~\sf sen(x)=\dfrac{1}{2}\implies x=arc~sen\bigg(\dfrac{1}{2}\bigg)\\\sf x=\dfrac{5\pi}{6}\\\tt a)~\sf cos(x)=cos\bigg(\dfrac{5\pi}{6}\bigg)=-cos\bigg(\dfrac{\pi}{6}\bigg)\\\sf cos(x)=-\dfrac{\sqrt{3}}{2}\\\tt b)~\sf sec(x)=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}\\\tt c)~\sf cossec(x)=2\\\tt d)~\sf tg(x)=tg\bigg(\dfrac{5\pi}{6}\bigg)=-tg\bigg(\dfrac{\pi}{6}\bigg)=-\dfrac{\sqrt{3}}{3}\\\tt e)~\sf cotg(x)=-\dfrac{3}{\sqrt{3}}=-\dfrac{3\sqrt{3}}{3}=-\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 4)~\sf sen(x)=\dfrac{1}{3}\implies sen^2(x)=\dfrac{1}{9}\\\sf cossec^2(x)=9\\\sf cotg^2(x)=cossec^2(x)-1\\\sf cotg^2(x)=9-1\\\sf cotg^2(x)=8\\\sf cotg(x)=\sqrt{8}\\\sf cotg(x)=2\sqrt{2}\\\sf\dfrac{\diagup\!\!\!\!\!sen(x)\cdot cos(x)}{\diagup\!\!\!\!sen^2(x)}=\dfrac{cos(x)}{sen(x)}\\\sf=cotg(x)=2\sqrt{2}\end{array}}$