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

1 Answer
A. S. Adikesavan
Jul 31, 2018

#- 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 )#.

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