How do you convert ${(r + 1)}^{2} = θ + csc θ$ $(r+1)^2= theta + csctheta$ to Cartesian form?

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How do you convert ${(r + 1)}^{2} = θ + csc θ$ $(r+1)^2= theta + csctheta$ to Cartesian form?

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Answer 1 · 2016-11-08T04:29:37

$r = \sqrt{x^{2} + y^{2}}$ $r=sqrt(x^2+y^2)$
$θ = {tan}^{- 1} (\frac{y}{x})$ $theta=tan^-1(y/x)$
${(\sqrt{x^{2} + y^{2}} + 1)}^{2} = {tan}^{- 1} (\frac{y}{x}) + \frac{y}{\sqrt{y^{2} + x^{2}}}$ $(sqrt(x^2+y^2)+1)^2=tan^-1(y/x)+y/sqrt(y^2+x^2)$

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How do you convert ${(r + 1)}^{2} = θ + csc θ$ $(r+1)^2= theta + csctheta$ to Cartesian form?

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How do you convert (r+1)^2= theta + csctheta (r+1)2=θ+cscθ (r+1)^2= theta + csctheta to Cartesian form?

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How do you convert ${(r + 1)}^{2} = θ + csc θ$ $(r+1)^2= theta + csctheta$ to Cartesian form?