Given $\theta \in [0,\pi)$, $\phi \in [0,2\pi)$, and $F$:
$F : \mathbb{R} \rightarrow \mathbb{R} \rightarrow \mathbb{R}^3,\; F(\theta, \phi) = \left[\cos(\phi)\sin(\theta),\ \sin(\phi)\sin(\theta),\ \cos(\theta) \right]$,
What are the formulas:
$G, H: \mathbb{R} \rightarrow \mathbb{R} \rightarrow \mathbb{R}^3 \\ G(\theta, \phi) = ? \\ H(\theta, \phi) = ?$,
such that
$F(\theta, \phi) \perp G(\theta, \phi) \perp H(\theta, \phi) \quad \forall \ \theta \in [0,\pi), \, \phi \in [0,2\pi).$
For a sphere, its tangential vectors satisfy the orthogonality. So we find
$$F_{\theta}=\frac{dF}{d\theta}=(\cos \phi\cos\theta, \sin\phi\cos\theta, -\sin\theta), F_{\phi}=\frac{dF}{d\phi}=(-\sin\phi\sin\theta, \cos\phi\sin\theta, 0)$$
You can see $F_{\theta}$ satisfies the conditions for $G$. But $F_{\phi}$ is not $(0,1,0)$ at $(0,0)$. To get the second component to equal $1$ when $\theta=0$ and $\phi=0$, we change the second component of $F_{\phi}$ to $\cos\phi\cos\theta$. Now to maintain orthogonality, which means dot product equals zero, we have to change the first component to $-\sin\phi\cos\theta$. This gives $$H=(-\sin\phi\cos\theta,\cos\phi\cos\theta,0)$$