What is the Cartesian form of rtheta+r = 2costheta-cottheta*sectheta rθ+r=2cosθ−cotθ⋅secθ?

1 Answer
Shiva Prakash M V
Feb 18, 2018

"Cartesian form of"Cartesian form of
rtheta+r=2costheta-cotthetaxxsecthetarθ+r=2cosθ−cotθ×secθ
" is given by " is given by
(sqrt(x^2+y^2))(1+tan^-1(y/x))-(sqrt(x^2+y^2))/y(√x2+y2)(1+tan−1(yx))−√x2+y2y

Explanation:

rtheta+r=2costheta-cotthetaxxsecthetarθ+r=2cosθ−cotθ×secθ

x=rcosthetax=rcosθ

y=rsinthetay=rsinθ

r=sqrt(x^2+y^2)r=√x2+y2

theta=tan^-1(y/x)θ=tan−1(yx)

tantheta=y/xtanθ=yx

costheta=x/r=x/sqrt(x^2+y^2)cosθ=xr=x√x2+y2

cottheta=1/tantheta=1/(y/x)=x/ycotθ=1tanθ=1yx=xy
sectheta=1/costheta=1/(x/sqrt(x^2+y^2))=(sqrt(x^2+y^2))/xsecθ=1cosθ=1x√x2+y2=√x2+y2x

Thus,

rtheta+r=2costheta-cotthetaxxsectheta " becomes"rθ+r=2cosθ−cotθ×secθ 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(√x2+y2)(tan−1(yx))+(√x2+y2)=2×x√x2+y2−xy×√x2+y2x
#

Simplifying

(sqrt(x^2+y^2))(1+tan^-1(y/x))-(sqrt(x^2+y^2))/y(√x2+y2)(1+tan−1(yx))−√x2+y2y

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