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

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Answer 1 · 2016-11-11T17:16:35

I would rewrite and find #C# first.

#secx/cscy = siny/cosx = C#

Given that the point #(pi/3, pi/3)# lies on the graph, we find #C = sqrt3#

Therefore,

#siny = sqrt3 cosx#.

Differentiating gets us

#cosy dy/dx = -sqrt3 sin x#

#dy/dx = -sqrt3 sinx/cosy#.

At #(pi/3, pi/3)# then, we have

#dy/dx = -sqrt3 sin(pi/3)/cos(pi/3) = -3#