What is the Cartesian form of #rtheta+r = 2costheta-cottheta*sectheta #?

1 Answer
Shiva Prakash M V
Feb 18, 2018

#"Cartesian form of"#
#rtheta+r=2costheta-cotthetaxxsectheta#
# " is given by "#
#(sqrt(x^2+y^2))(1+tan^-1(y/x))-(sqrt(x^2+y^2))/y#

Explanation:

#rtheta+r=2costheta-cotthetaxxsectheta#

#x=rcostheta#

#y=rsintheta#

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

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

#tantheta=y/x#

#costheta=x/r=x/sqrt(x^2+y^2)#

#cottheta=1/tantheta=1/(y/x)=x/y#
#sectheta=1/costheta=1/(x/sqrt(x^2+y^2))=(sqrt(x^2+y^2))/x#

Thus,

#rtheta+r=2costheta-cotthetaxxsectheta " becomes"#

#(sqrt(x^2+y^2))(tan^-1(y/x))+(sqrt(x^2+y^2))=2xxx/sqrt(x^2+y^2)-x/yxx(sqrt(x^2+y^2))/x#
#

Simplifying

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

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