Metric on the embedded hyperboloid in five-dimensional Minkowski space I have an explicit parametrisation of a one-sheeted hyperboloid in the five-dimensional Minkowski space, namely:
$$Z_{0}=-l \cdot \cot(\tau)$$
$$Z_{a}=\frac{l}{\sin(\tau)}\omega_{a}, \quad a=1,\ldots,4$$
Where the $\omega_{a}$ represent the coordinates that embed a unit three sphere into $\Bbb R^4$ and satisfy $\omega_{a} \omega_{a}=1$. In the ambient space we have the metric:
$$ds^2=-dZ_{0}^2+dZ_{1}^2+dZ_{2}^2+dZ_{3}^2+dZ_{4}^2.$$
Now I am trying to verify that for the coordinates above the metric takes the form:

$$ds^2=\frac{l^2}{\sin^2(\tau)}(-d \tau^2 +d\Omega_{3}^2)$$

Where $d\Omega_{3}^2$ denotes the metric of the unit three-sphere.
I am extremly thankful for every answer, comment or idea!
Thank you all in advance!

Let us set $l=1$. One can reinsert the right power of $l$ at the end, as it is nothing but a scaling constant.
Let me first correct that the problem actually concerns the hyperboloid with one sheet $M$ of all unit spacelike vectors in Minkowski $5$-dimensional spacetime, namely the vectors $Z^a$ satisfying $Z^a Z_a = 1$, where I am implicitly using Einstein's summation convention over indices $a$ going from $0$ to $4$, and I have used the Minkowski metric to lower the index of the second $Z$.
The problem is essentially a change of coordinate problem. Let us delve into the details. We have
$$dZ_0 = - \csc^2(\tau)d\tau,$$ which gives $$dZ_0^2 = \csc^4(\tau)d\tau^2.$$
Similarly,
$$dZ_i = -\csc(\tau)\cot(\tau) \omega_i d\tau + \csc(\tau) d\omega_i,$$ which, upon squaring, gives
$$dZ_i^2 = \csc^2(\tau)\cot^2(\tau)\omega_i^2d\tau^2 + \csc^2(\tau)d\omega_i^2 
- 2 \csc^2(\tau)\cot(\tau) \omega_i d\omega_i d\tau.$$ Note that $\sum_{i=1}^4 \omega_i^2 = 1$, which implies, upon differentiating, that $\sum_{i=1}^4 \omega_i d\omega_i = 0$. This implies that after taking a summation for $i$ going from $1$ to $4$, the last term in the formula for $dZ_i^2$ above vanishes.
Hence the metric on the $4$-dimensional hyperboloid with one sheet $M$ becomes in the new coordinates:
$$ ds^2 = -(\csc^4(\tau) - \csc^2(\tau)\cot^2(\tau))d\tau^2 + \csc^2(\tau) 
\sum_{i=1}^4 d\omega_i^2.$$
After factoring out $\csc^2(\tau)$ and using the trigonometric identity $\csc^2(\tau)-\cot^2(\tau) = 1$, one gets the sought-after expression, albeit with $l$ set to $1$ (see my remark at the beginning of this post).
