Let us convert #r=2csctheta+5# to Cartesian coordinates by using #rcostheta=x#, #rsintheta=y# and #r^2=x^2+y^2#. We can write #r=2csctheta+5# as
#rsintheta=2+5sintheta# or #y=2+y/sqrt(x^2+y^2)#
i.e. #y/sqrt(x^2+y^2)=y-2#
or #y^2/(x^2+y^2)=(y-2)^2#
or #(x^2+y^2)(y-2)^2=y^2#
Now horizontal tangents are there where #(dy)/(dx)=0# and hence finding #(dy)/(dx)# using implicit differentiiation, we get
#2(x^2+y^2)(y-2)(dy)/(dx)+(y-2)^2(2x+2y(dy)/(dx))=2y(dy)/(dx)#
or #(dy)/(dx)[2(x^2+y^2)(y-2)+2y(y-2)^2-2y]=-2x(y-2)^2#
and hence #(dy)/(dx)=0# when either #y=2# or #x=0#
Now #y=2# means #rsintheta=2# but as equation can be written as #rsintheta=2+5sintheta#, this means when #y=2#, we have #sintheta=0# or #theta=0#. But as function #r=2csctheta+5# s not defined at #theta=0#, we ignore this.
#x=0# means #rcostheta=0# or #costheta=0#, i.e. #theta=pi/2# or #(3pi)/2# and hence points of horizontal tendency are #theta=pi/2# and #(3pi)/2#
graph{(x^2+y^2)(y-2)^2=y^2 [-10, 10, -5, 5]}
