Respuesta :
[tex]\bf cot(\theta)=\cfrac{adjacent}{opposite}\qquad \qquad sec(\theta)=\cfrac{hypotenuse=c}{adjacent=a}\\\\
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sec(\theta)=-2\iff sec(\theta)=-\cfrac{2}{1} \cfrac{\leftarrow hypotenuse}{\leftarrow adjacent}[/tex]
so hmmm let's recall, the hypotenuse, or radius, is never negative, since it's just a unit of the radius, so from -2/1 the negative fellow has to be the denominator, namely, the adjacent side, so a = -1, c = 2
bear also in mind, the angle is in the III quadrant, and on that quadrant, the sign for the x-coordinate is negative, so is -1
so.. what's the opposite side? or b? let's use the pythagorean theorem then
[tex]\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{(2)^2-(-1)^2}=b \\\\\\ \pm\sqrt{3}=b[/tex]
so.. which is it? the +/-? well, we're on the III quadrant, on there both the "x" and "y" coordinates are both negative, so "b" has to also be negative, thus -√(3)=b
[tex]\bf cot(\theta)=\cfrac{adjacent}{opposite}\implies cot(\theta)=\cfrac{a}{b}\implies cot(\theta)=\cfrac{-1}{-\sqrt{3}} \\\\\\ cot(\theta)=\cfrac{1}{\sqrt{3}}\impliedby \textit{now, let's rationalize that}\\\\ -------------------------------\\\\ \cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \boxed{\cfrac{\sqrt{3}}{3}}[/tex]
so hmmm let's recall, the hypotenuse, or radius, is never negative, since it's just a unit of the radius, so from -2/1 the negative fellow has to be the denominator, namely, the adjacent side, so a = -1, c = 2
bear also in mind, the angle is in the III quadrant, and on that quadrant, the sign for the x-coordinate is negative, so is -1
so.. what's the opposite side? or b? let's use the pythagorean theorem then
[tex]\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{(2)^2-(-1)^2}=b \\\\\\ \pm\sqrt{3}=b[/tex]
so.. which is it? the +/-? well, we're on the III quadrant, on there both the "x" and "y" coordinates are both negative, so "b" has to also be negative, thus -√(3)=b
[tex]\bf cot(\theta)=\cfrac{adjacent}{opposite}\implies cot(\theta)=\cfrac{a}{b}\implies cot(\theta)=\cfrac{-1}{-\sqrt{3}} \\\\\\ cot(\theta)=\cfrac{1}{\sqrt{3}}\impliedby \textit{now, let's rationalize that}\\\\ -------------------------------\\\\ \cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \boxed{\cfrac{\sqrt{3}}{3}}[/tex]
