# What is the slope of the tangent line of secx-cscy= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Jul 31, 2018

$- 3 \sqrt{3}$

#### Explanation:

As a point of the curve is $T \left(\frac{\pi}{3} , \frac{\pi}{3}\right)$,

$C = \sec \left(\frac{1}{3} \pi\right) - \csc \left(\frac{1}{3} \pi\right) = 2 - \frac{2}{\sqrt{3}}$, So,

$\sec x - \csc y = 2 \left(1 - \frac{1}{3} \sqrt{3}\right)$. And so,

$\sec x \tan x - \left(- \csc y \cot y\right) y ' = 0$, giving

y' at T = - ( sec (1/3pi ) tan ( 1/3pi))/( csc ( 1/3pi )(cot ( 1/3pi ))

= - (( 2 ) ( sqrt 3 ))/(( 2/sqrt3 )(1/sqrt3 )

$= - 3 \sqrt{3}$

Graph, with tangent:
graph{(1/cos x- 1/ sin y - 2 ( 1- 1/3 sqrt3 ))(y-1/3pi+3sqrt3(x-1/3pi))(y-1/3pi-3sqrt3(x+1/3pi))((x-1/3pi)^2+(y-1/3pi)^2-0.025)((x+1/3pi)^2+(y-1/3pi)^2-0.025)=0}

Indeed, vivid.

Did you observe that I have shown also

the tangent at $T ' \left(- \frac{1}{3} \pi , \frac{1}{3} \pi\right)$.