Derivation of the graph of $r=\frac{1}{\sin(\theta)}$

Question

When playing around with desmos, I found a very interesting function: $r=\frac{1}{\sin(\theta)}$

The graph of this function is a straight line with constant value $y=1$

I tried to prove this, however I failed:

Assuming $r=\sqrt {x^2+y^2}$ and $\theta = \arctan(\frac{y}{x})$ hold, we get:

$\sqrt {x^2+y^2} = \frac{1}{\sin(\arctan(\frac{y}{x}))}$

$\sqrt {x^2+y^2} = \frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$

$\sqrt {x^2+y^2} = \frac{1+\frac{y^2}{x^2}}{\frac{y}{x}}$ |$*\frac{y}{x}$

$\sqrt {\frac{y^2}{x^2}(x^2+y^2)} = 1+\frac{y^2}{x^2}$

$\sqrt {y^2 + \frac{y^4}{x^2}} = 1+\frac{y^2}{x^2}$ |$^2$

$y^2+\frac{y^4}{x^2} = 1+2\frac{y^2}{x^2}+\frac{y^4}{x^4}$ |$*x^4$

$x^4 y^2 + x^2 y^4 = x^4 + 2x^2 y^2 + y^4$

$(x^2 y^2)(x^2+y^2)=(x^2+y^2)^2$ |$:(x^2+y^2)$

$x^2 y^2 = x^2+y^2$ |$-y^2$

$x^2 y^2 - y^2 = x^2$

$(x^2 - 1) y^2 = x^2$ |$:(x^2-1)$

$y^2 = \frac{x^2}{x^2-1}$

$y = \sqrt{\frac{x^2}{x^2-1}}$

The graph of this function is obviously not a straight line with $y=1$.

Where did my calculation go wrong and what would be the actual derivation?

Answer 1

I found the error myself! It lies within my very first step:

$\frac{1}{\sin(\arctan(\frac{y}{x}))}$ is not equal to $\frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$ but rather to $\frac{1}{\frac{\frac{y}{x}}{\sqrt{1+\frac{y^2}{x^2}}}}$

From there, the derivation is straightforward:

$\sqrt{x^2+y^2} = \frac{\sqrt{1+\frac{y^2}{x^2}}}{\frac{y}{x}}$

$\sqrt {y^2 + \frac{y^4}{x^2}} = \sqrt{1+\frac{y^2}{x^2}}$

$y^2 + \frac{y^4}{x^2} = 1+\frac{y^2}{x^2}$

$y^2 (1+\frac{y^2}{x^2}) = 1+\frac{y^2}{x^2}$

$y^2=1$

$y=\pm1$

Answer 2

A correct derivation is much simpler from the Cartesian-to-polar transformation, whereas you used the polar-to-Cartesian transformation and overcomplicated things: $$r=\frac1{\sin\theta}\implies r\sin\theta=y=1$$