How do you convert # (r+1)^2= 2r-theta + sin theta costheta # to Cartesian form?

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Answer 1 · 2018-06-27T01:23:07

#x^2+y^2+1=(xy)/(x^2+y^2)-tan^-1(y/x)#

Expand and Simplify

#r^2+1=-theta+sin(theta)cos(theta)#

Recall #x^2+y^2=r^2,##x=rcostheta#, #y=rsintheta#

Therfore #y/x=tan(theta)# this means #tan^-1(y/x)=theta#

Trickery

#r^2+1=-theta+((rsintheta)(rcostheta))/r^2#

Substitute

#x^2+y^2+1=(xy)/(x^2+y^2)-tan^-1(y/x)#