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

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

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Answer 1 · 2016-06-29T17:17:46

$\frac{x}{y} = tan [x^{2} + y^{2} + \sqrt{x^{2} + y^{2}} (2 - \frac{1}{x}) + 1]$ $x/y=tan[x^2+y^2+sqrt(x^2+y^2)(2-1/x)+1]$

Explanation:

When we convert polar coordinates $(r, θ)$ $(r,theta)$ to Cartesian coordinates, the relation is $x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ and hence $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $θ = {tan}^{- 1} (\frac{x}{y})$ $theta=tan^(-1)(x/y)$ .

Hence ${(r + 1)}^{2} = θ + sec θ$ $(r+1)^2=theta+sectheta$ can be written as

$r^{2} + 2 r + 1 = θ + sec θ$ $r^2+2r+1=theta+sectheta$ or

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

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

$\frac{x}{y} = tan [x^{2} + y^{2} + \sqrt{x^{2} + y^{2}} (2 - \frac{1}{x}) + 1]$ $x/y=tan[x^2+y^2+sqrt(x^2+y^2)(2-1/x)+1]$

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

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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Impact of this question

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