Your equation is equivalent to [itex]\rho[/itex] = cos([itex]\phi[/itex])/(1 - cos2([itex]\phi[/itex]).
Your relevant equations show the converstion from spherical to Cartesian coordinates. Do you know the conversions going in the other direction?
One trick that will be helpful is to multiply both sides by rho. This potentially adds a point (rho = 0) that might not be a solution of your original equation, so you should check whether this is already a solution of your equation.
As an intermediate step, look at what I have in my previous post. Use trig identities to get to what I showed, then go from there.
Look at the equations for x, y, and z. Note that they're all of the form rho times some combination of sines and cosines. So the first thing you should do is rewrite everything in terms of sines and cosines. Next, rearrange terms so the trig functions multiply rho. If needed, do as Mark suggested, and multiply both sides by rho. Then try to identify what cartesian coordinates the various products are equal to.
You're missing a factor of cos(phi) in the numerator on the right side.
Try bringing the (1+cos^2(phi)) to the other side.
OK. I didn't catch what you were doing.
Don't you mean (1 - cos^2(phi))?
It's probably helpful not to change the rho^2 on the left side just yet.
Let us decide whether nothing good comes out of it. Show us what you tried.
You have rho^2 = z/(1 - cos^2(phi))
What do you get when you multiply both sides by 1 - cos^2(phi)?
Yes, I did. Thanks for catching that.
That's a very nice thing to say! Thank you!
csc(x) = 1/sin(x), cot(x) = cos(x)/sin(x)
Are those the ones you mean?
Notice to Homework helpers. Please be sure to use the quote function to make a copy of the poster's problem.
Separate names with a comma.