Respuesta :
[tex]\bf y(x) = 5csc(x)-5cot(x)\implies y\left( \frac{\pi }{3} \right)= 5csc\left( \frac{\pi }{3} \right)-5cot\left( \frac{\pi }{3} \right) \\\\\\ y\left( \frac{\pi }{3} \right) = 5\cdot \cfrac{2}{\sqrt{3}}-5\cdot \cfrac{1}{\sqrt{3}}\implies y\left( \frac{\pi }{3} \right)=\cfrac{10}{\sqrt{3}}-\cfrac{5}{\sqrt{3}}\implies y\left( \frac{\pi }{3} \right)=\cfrac{5}{\sqrt{3}}[/tex]
[tex]\bf y\left( \frac{\pi }{3} \right)=\cfrac{5\sqrt{3}}{\sqrt{3}\sqrt{3}}\implies y\left( \frac{\pi }{3} \right)=\cfrac{5\sqrt{3}}{3}~\hfill \stackrel{\textit{this gives us the point of }}{\left( \frac{\pi }{3}~~,~~ \frac{5\sqrt{3}}{3}\right)} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dy}{dx}=-5csc(x)cot(x)+5csc^2(x) \right|_{x = \frac{\pi }{3}}\implies -5csc\left( \frac{\pi }{3} \right)cot\left( \frac{\pi }{3} \right)+5csc^2\left( \frac{\pi }{3} \right)[/tex]
[tex]\bf -5\cdot \cfrac{2}{\sqrt{3}}\cdot \cfrac{1}{\sqrt{3}}+5\left( \cfrac{2}{\sqrt{3}} \right)^2\implies \cfrac{-10}{3}+5\cdot \cfrac{4}{3}\implies \cfrac{-10}{3}+\cfrac{20}{3}\implies \stackrel{slope}{\cfrac{10}{3}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5\sqrt{3}}{3}}=\stackrel{m}{\cfrac{10}{3}}\left(x-\stackrel{x_1}{\cfrac{\pi }{3}}\right)[/tex]
[tex]\bf y-\cfrac{5\sqrt{3}}{3}=\cfrac{10}{3}x-\cfrac{10\pi }{9}\implies y=\cfrac{10}{3}x-\cfrac{10\pi }{9}+\cfrac{5\sqrt{3}}{3} \\\\\\ y=\cfrac{10}{3}x+\cfrac{-10\pi +15\sqrt{3}}{9}\implies y=\cfrac{10}{3}x-\cfrac{10\pi -15\sqrt{3}}{9}[/tex]
