Let $\Delta =$$\begin{vmatrix} sin\theta cos \phi & sin\theta sin\phi & cos\theta \\ cos\theta cos\phi & cos\theta sin\phi & -sin\theta \\ -sin\theta sin\phi & sin\theta cos\phi & 0\end{vmatrix}$, then

$\triangle = \begin{vmatrix} \sin \theta . \cos \phi & \sin \theta . \sin \phi & \cos \theta \\ \cos \theta . \cos \phi & \cos \theta . \sin \theta & -\sin \theta \\ -\sin \theta. \sin \theta & \sin \theta . \cos \phi & 0 \end{vmatrix}$

$= \sin \theta . \sin \phi (\sin^{2} \theta . \cos \theta) - \cos \theta (- \sin \theta . \cos \theta . \cos \theta)- \sin \theta . \sin \phi (- \sin^{2} \theta. \sin \phi - \cos^{2} \theta- \sin \theta)$

$=\sin^{3} \theta . \sin \phi. \cos \phi+ \cos^{2} \theta. \sin \theta. \cos^{2} \phi+ \sin^{3} \theta. \sin^{2} \phi + \sin \theta . \cos^{2} \theta . \sin^{2} \phi$

So, We can say that $\triangle$ is depend on both

$\theta$ and $\phi$