Re-writing a positional vector in terms of trigonometric functions

Question

A particle of mass $ m $ and position $ \mathbf{r} = x \cdot \mathbf{i} + y \cdot \mathbf{j} + z \cdot \mathbf{k} $ is constrained to a smooth surface described by

$$ x^{2} + z^{2} = 1 + y^{2}, $$ where $ - \pi \leq y \leq \pi $.

Can we write $ \mathbf{r} $ in terms of trigonometric functions?

Answer 1

$$x=\sec{\theta} \cos{\phi}$$ $$y=\tan{\theta} $$ $$z=\sec{\theta} \sin{\phi} $$

$\phi \in [0,2 \pi)$, $\theta \in [-\arctan{\pi},\arctan{\pi}]$.

Answer 2

You can write $$ x= \rho \cos \theta\\ z= \rho \sin \theta\\ y= \pm\sqrt{\rho^2-1} $$