Question

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

Answer 1

`- 3 sqrt 3`

Explanation:

As a point of the curve is `T ( pi/3, pi/3 )`,

`C = sec (1/3pi) - csc (1/3pi) = 2 - 2/sqrt3`, So,

`sec x - csc y = 2 ( 1- 1/3sqrt3 )`. And so,

`secx tanx - (- csc y cot y )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'( -1/3pi, 1/3pi )`.