Question

Find an equation of the tangent line to y=5csc(x)−5cotx at x=π/3.

(Use symbolic notation and fractions where needed.)


Y=?

Answer 1

`\bf y(x) = 5csc(x)-5cot(x)\implies y\left( (\pi )/(3) \right)= 5csc\left( (\pi )/(3) \right)-5cot\left( (\pi )/(3) \right) \\\\\\ y\left( (\pi )/(3) \right) = 5\cdot \cfrac{2}{√(3)}-5\cdot \cfrac{1}{√(3)}\implies y\left( (\pi )/(3) \right)=\cfrac{10}{√(3)}-\cfrac{5}{√(3)}\implies y\left( (\pi )/(3) \right)=\cfrac{5}{√(3)}`

`\bf y\left( (\pi )/(3) \right)=\cfrac{5√(3)}{√(3)√(3)}\implies y\left( (\pi )/(3) \right)=\cfrac{5√(3)}{3}~\hfill \stackrel{\textit{this gives us the point of }}{\left( (\pi )/(3)~~,~~ (5√(3))/(3)\right)} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dy}{dx}=-5csc(x)cot(x)+5csc^2(x) \right|_{x = (\pi )/(3)}\implies -5csc\left( (\pi )/(3) \right)cot\left( (\pi )/(3) \right)+5csc^2\left( (\pi )/(3) \right)`

`\bf -5\cdot \cfrac{2}{√(3)}\cdot \cfrac{1}{√(3)}+5\left( \cfrac{2}{√(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}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√(3)}{3}}=\stackrel{m}{\cfrac{10}{3}}\left(x-\stackrel{x_1}{\cfrac{\pi }{3}}\right)`

`\bf y-\cfrac{5√(3)}{3}=\cfrac{10}{3}x-\cfrac{10\pi }{9}\implies y=\cfrac{10}{3}x-\cfrac{10\pi }{9}+\cfrac{5√(3)}{3} \\\\\\ y=\cfrac{10}{3}x+\cfrac{-10\pi +15√(3)}{9}\implies y=\cfrac{10}{3}x-\cfrac{10\pi -15√(3)}{9}`