Let Delta = begin{vmatrix} sin theta cos phi | vedclass

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Question

(B, D) Expanding the determinant $\Delta$ along the third column: $\Delta = \cos 	heta \begin{vmatrix} \cos 	heta \cos \phi & \cos 	heta \sin \phi

Vedclass · Accepted Answer

$\left(\frac{d \Delta}{d 	heta}ight)_{	heta = \frac{\pi}{2}} = 0$

Vedclass · Answer

(B, D) Expanding the determinant $\Delta$ along the third column: $\Delta = \cos 	heta \begin{vmatrix} \cos 	heta \cos \phi & \cos 	heta \sin \phi \ -\sin 	heta \sin \phi & \sin 	heta \cos \phi \end{vmatrix} - (- \sin 	heta) \begin{vmatrix} \sin 	heta \cos \phi & \sin 	heta \sin \phi \ -\sin 	heta \sin \phi & \sin 	heta \cos \phi \end{vmatrix} + 0$ $\Delta = \cos 	heta [(\cos 	heta \cos \phi)(\sin 	heta \cos \phi) - (\cos 	heta \sin \phi)(-\sin 	heta \sin \phi)] + \sin 	heta [(\sin 	heta \cos \phi)(\sin 	heta \cos \phi) - (\sin 	heta \sin \phi)(-\sin 	heta \sin \phi)]$ $\Delta = \cos 	heta [\sin 	heta \cos 	heta \cos^2 \phi + \sin 	heta \cos 	heta \sin^2 \phi] + \sin 	heta [\sin^2 	heta \cos^2 \phi + \sin^2 	heta \sin^2 \phi]$ $\Delta = \cos 	heta [\sin 	heta \cos 	heta (\cos^2 \phi + \sin^2 \phi)] + \sin 	heta [\sin^2 	heta (\cos^2 \phi + \sin^2 \phi)]$ $\Delta = \cos 	heta (\sin 	heta \cos 	heta) + \sin 	heta (\sin^2 	heta) = \sin 	heta \cos^2 	heta + \sin^3 	heta = \sin 	heta (\cos^2 	heta + \sin^2 	heta) = \sin 	heta$ Since $\Delta = \sin 	heta$, it is independent of $\phi$. Also, $\frac{d \Delta}{d 	heta} = \cos 	heta$. At $	heta = \frac{\pi}{2}$, $\frac{d \Delta}{d 	heta} = \cos \frac{\pi}{2} = 0$. Thus, both options $B$ and $D$ are correct.

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:

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