Question

sendo cos=-2/3 , com pi/2 < x < pi , determine os valores de tanx e cossec x .......????

Answer 1

$\cos x=-\dfrac{2}{3},\;\;\dfrac{\pi}{2}<x<\pi$


Utilizando a Relação Fundamental da Trigonometria, encontramos o seno de $x$:

$\cos^{2} x+\mathrm{sen^{2}\,}x=1\\ \\ \\ \mathrm{sen^{2}\,}x=1-\cos^{2}x\\ \\ \mathrm{sen^{2}\,}x=1-\left(-\dfrac{2}{3} \right )^{2}\\ \\ \mathrm{sen^{2}\,}x=1-\dfrac{4}{9}\\ \\ \mathrm{sen^{2}\,}x=\dfrac{9-4}{9}\\ \\ \mathrm{sen^{2}\,}x=\dfrac{5}{9}\\ \\ \mathrm{sen\,}x=\pm \sqrt{\dfrac{5}{9}}\\ \\ \mathrm{sen\,}x=\pm \dfrac{\sqrt{5}}{3}$


Como $x$ é um arco do segundo quadrante, o seu seno é positivo. Logo,

$\mathrm{sen\,}x=\dfrac{\sqrt{5}}{3}$


$\bullet\;\;\tan x=\dfrac{\mathrm{sen\,}x}{\cos x}\\ \\ \\ \tan x=\dfrac{\sqrt{5}/3}{-2/3}\\ \\ \tan x=\dfrac{\sqrt{5}}{\diagup\!\!\!\! 3}\cdot \left(-\dfrac{\diagup\!\!\!\! 3}{2} \right )\\ \\ \tan x=-\dfrac{\sqrt{5}}{2}$


$\bullet\;\;\mathrm{cossec\,} x=\dfrac{1}{\mathrm{sen\,}x}\\ \\ \\ \mathrm{cossec\,} x=\dfrac{1}{\sqrt{5}/3}\\ \\ \mathrm{cossec\,} x=\dfrac{3}{\sqrt{5}}\\ \\ \mathrm{cossec\,} x=\dfrac{3\cdot \sqrt{5}}{\sqrt{5}\cdot \sqrt{5}}\\ \\ \mathrm{cossec\,} x=\dfrac{3\sqrt{5}}{5}$