What is the Cartesian form of #(r+3)^2 = sin^3theta-sec^2theta #?

1 Answer
Feb 19, 2018

#(r+3)^2=sin^3theta-sec^2theta# when expressed in catesian form becomes

#(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)#
which can be simplified as required

Explanation:

#x=rcostheta#

#y=rsintheta#

#r=sqrt(x^2+y^2)#
#r+3=sqrt(x^2+y^2)+3=3+sqrt(x^2+y^2)#
#(r+3)^2=(3+sqrt(x^2+y^2))^2#

#sin^3theta=(sintheta)^3#
#sintheta=y/r=y/sqrt(x^2+y^2)#
#sin^3theta=(y/sqrt(x^2+y^2))^3#

#sec^2theta=1+tan^2theta#
#tan^2theta=(tantheta)^2#
#theta=tan^-1(y/x)#
#tantheta=y/x#
#tan^2theta=(y/x)^2#
#sec^2theta=1+(y/x)^2#
Now, we have
#(r+3)^2=(3+sqrt(x^2+y^2))^2#
#sin^3theta=(y/sqrt(x^2+y^2))^3#
#sec^2theta=1+(y/x)^2#

Thus,
#(r+3)^2=sin^3theta-sec^2theta# when expressed in catesian form becomes

#(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)#
which can be simplified as required