Calc 3 Help- Polar to rectangular

Hello im having trouble converting this equation to rectangular.

r = 1/ 1+ cot^2 Theta

Any help would be great

It helps to recall that cot²Θ + 1 = csc²Θ.

r = 1/csc²Θ = sin²Θ ==>

r^3 = r² sin²Θ ==> (x² + y²)^(3/2) = y².

You can make it a bit prettier by squaring both sides and collecting some terms.

(x² + y²)^3 = y^4 ==> x^6 + y^6 + 3x^4y² + 3x²y^4 - y^4 = 0.

We can start out by re-writing 1 + cot^2θ as csc^2θ. This gives:
r = 1/(1 + cot^2θ)
==> r = 1/csc^2θ
==> r = sin^2θ, since 1/cscθ = sinθ.

Multiplying both sides by r^2 gives:
r^3 = r^2*sin^2θ
==> r^3 = (r*sinθ)^2.

Then, squaring both sides gives:
r^6 = (r*sinθ)^4
==> (r^2)^3 = (r*sinθ)^4
==> (x^2 + y^2)^3 = y^4, by substituting in r^2 = x^2 + y^2 and r*sinθ = y.

I hope this helps!

NB: I simplified theta as t.

r = 1/(1+cot^2(t))
= 1/(1+(cos^2(t)/sin^2(t)))
= sin^2(t)/(sin^2(t) + cos^2(t))
= sin^2(t)

multiply each side with r^2.
r^3 = (r^2)(sin^2(t)) = y^2 (since y = rsin(t))

now substitute r.
(x^2 + y^2)^(3/2) = y^2


Hooray!