How do you simplify `csc^2(θ) / (1 + tan^2(θ) )`?

Let's see, we have:

`(csc^2 theta)/(1+tan^2 theta)`

In this case, I would probably start by rewriting everything in terms of sines and cosines, since we know that
`tan theta = sin theta/cos theta`
and
`csc theta = 1/sin theta`

So we now have:

`= (1/sin^2 theta) / (1+ sin^2 theta /cos^2 theta)`

Here, I notice that the denominator can be simplified if I would multiply it by `cos^2 theta`. Here is what I mean. The denominator is:
`1+(sin^2 theta)/(cos^2 theta)`
and if I multiply `cos^2 theta`, we get:
`(1+(sin^2 theta)/(cos^2 theta) )* cos^2 theta`
Opening the parentheses, this becomes:

`cos^2 theta + sin^2 theta`
which is our famous identity and is `=1`.

Yes. The denominator becomes 1 if I would multiply by `cos^2 theta`.
So let's do the following:

`= (1/sin^2 theta) / (1+ sin^2 theta /cos^2 theta) * (cos^2 theta)/(cos^2 theta)`

We can do this since any number over itself (except zero) is equal to 1.
So provided that `theta != pi/2` or `-pi/2`, we can multiply our equation by
`cos^2 theta / cos^2 theta = 1`

Now, we know the denominator becomes 1, and we are left with just the "top":

`= cos^2 theta / sin^2 theta`
`=cot^2 theta`

where `cot theta` is the cotangent of theta.

Use the trig identity: `1 + tan^2 x = sec^2 x`. We get:

`f(x) = (csc^2 x)/(sec^2 x) = (1/sin^2 x)/(1/cos^2 x) =`
`= (cos^2 x)/(sin^2 x) = cot^2 x`