Respuesta :
now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
[tex]\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}[/tex]
recall that
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad\qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \\\\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
therefore, let's just plug that on the remaining ones,
[tex]\bf sin(\theta)=\cfrac{-1}{6} \qquad\qquad cos(\theta)=\cfrac{\sqrt{35}}{6} \\\\\\ % tangent tan(\theta)=\cfrac{-1}{\sqrt{35}} \qquad \qquad % cotangent cot(\theta)=\cfrac{\sqrt{35}}{1} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}[/tex]
now, let's rationalize the denominator on tangent and secant,
[tex]\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}[/tex]
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
[tex]\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}[/tex]
recall that
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad\qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \\\\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
therefore, let's just plug that on the remaining ones,
[tex]\bf sin(\theta)=\cfrac{-1}{6} \qquad\qquad cos(\theta)=\cfrac{\sqrt{35}}{6} \\\\\\ % tangent tan(\theta)=\cfrac{-1}{\sqrt{35}} \qquad \qquad % cotangent cot(\theta)=\cfrac{\sqrt{35}}{1} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}[/tex]
now, let's rationalize the denominator on tangent and secant,
[tex]\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}[/tex]
