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Let $\Delta=\left|\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0\end{array}\right|$. Then
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$\Delta$ is independent of $\varphi$, $\left(\frac{d \Delta}{d \theta}\right)_{\theta-\frac{\pi}{2}}=0$
$\Delta=\left|\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0\end{array}\right|$
$\begin{aligned}
\mathrm{C}_1 \longrightarrow \mathrm{C}_1-\left(\frac{\cos \phi}{\sin \phi}\right) \mathrm{C}_2 \Rightarrow \Delta=\left|\begin{array}{ccc}
0 & \sin \theta \sin \phi & \cos \theta \\
0 & \cos \theta \sin \phi & -\sin \theta \\
-\frac{\sin \theta}{\sin \phi} & \sin \theta \cos \phi & \theta
\end{array}\right| \\
&=\frac{\sin \theta}{\sin \phi} \cdot \sin \phi\left(\cos ^2 \theta+\sin ^2 \theta\right)= \sin \theta \rightarrow \text { Independent of } \phi \\
\frac{\mathrm{d} \Delta}{\mathrm{d} \theta}=\left.\cos \theta \Rightarrow \frac{\mathrm{d} \Delta}{\mathrm{d} \theta}\right|_{\theta-\frac{\pi}{2}}=0
\end{aligned}$
$\begin{aligned}
\mathrm{C}_1 \longrightarrow \mathrm{C}_1-\left(\frac{\cos \phi}{\sin \phi}\right) \mathrm{C}_2 \Rightarrow \Delta=\left|\begin{array}{ccc}
0 & \sin \theta \sin \phi & \cos \theta \\
0 & \cos \theta \sin \phi & -\sin \theta \\
-\frac{\sin \theta}{\sin \phi} & \sin \theta \cos \phi & \theta
\end{array}\right| \\
&=\frac{\sin \theta}{\sin \phi} \cdot \sin \phi\left(\cos ^2 \theta+\sin ^2 \theta\right)= \sin \theta \rightarrow \text { Independent of } \phi \\
\frac{\mathrm{d} \Delta}{\mathrm{d} \theta}=\left.\cos \theta \Rightarrow \frac{\mathrm{d} \Delta}{\mathrm{d} \theta}\right|_{\theta-\frac{\pi}{2}}=0
\end{aligned}$
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