Question

A determinant is given as following:
$\Delta =\left| \begin{matrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0 \\\ \end{matrix} \right|$
Then,
A. $\Delta$ is independent of $\theta$
B. $\Delta$ is independent of $\phi$
C. $\Delta$ is constant.
D. ${{\left. \dfrac{d\Delta }{d\theta } \right|}_{\theta =\dfrac{\pi }{2}}}=0$

Answer 1

Hint: Find out the value of the determinant and check if the value is a constant or free from $\theta$ or free from $\phi$. Then differentiate the result with respect to $\theta$ and check if the result of the derivative at $\theta =\dfrac{\pi }{2}$ is zero or not.

“Complete step-by-step answer:”
The determinant is given as follows:
$\Delta =\left| \begin{matrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0 \\\ \end{matrix} \right|$
Now let us first find the value of the determinant.
$\Delta =\sin \theta \cos \phi \left| \begin{matrix} \cos \theta \sin \phi & -\sin \theta \\\ \sin \theta \cos \phi & 0 \\\ \end{matrix} \right|-\sin \theta \sin \phi \left| \begin{matrix} \cos \theta \cos \phi & -\sin \theta \\\ -\sin \theta \sin \phi & 0 \\\ \end{matrix} \right|+\cos \theta \left| \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\\ -\sin \theta \sin \phi & \sin \theta \cos \phi \\\ \end{matrix} \right|$ Now we will take the order of two determinants one by one and solve them.
At first we have,
$\left| \begin{matrix} \cos \theta \sin \phi & -\sin \theta \\\ \sin \theta \cos \phi & 0 \\\ \end{matrix} \right|=(\cos \theta \sin \phi \times 0)-(-\sin \theta )\times \sin \theta \cos \phi ={{\sin }^{2}}\theta \cos \phi$
Multiplying the above determinant by $\sin \theta \cos \phi$,$\sin \theta \cos \phi \left| \begin{matrix} \cos \theta \sin \phi & -\sin \theta \\\ \sin \theta \cos \phi & 0 \\\ \end{matrix} \right|={{\sin }^{2}}\theta \cos \phi \times \sin \theta \cos \phi ={{\sin }^{3}}\theta {{\cos }^{2}}\phi$
Let us take the second determinant.
$\left| \begin{matrix} \cos \theta \cos \phi & -\sin \theta \\\ -\sin \theta \sin \phi & 0 \\\ \end{matrix} \right|=(\cos \theta \cos \phi \times 0)-\left( -\sin \theta \right)\times \left( -\sin \theta \sin \phi \right)=-{{\sin }^{2}}\theta \sin \phi$
Multiplying the above determinant by $-\sin \theta \sin \phi$,
$-\sin \theta \sin \phi \left| \begin{matrix} \cos \theta \cos \phi & -\sin \theta \\\ -\sin \theta \sin \phi & 0 \\\ \end{matrix} \right|=(-{{\sin }^{2}}\theta \sin \phi )\times \left( -\sin \theta \sin \phi \right)={{\sin }^{3}}\theta {{\sin }^{2}}\phi$
And the third determinant,
$\left| \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\\ -\sin \theta \sin \phi & \sin \theta \cos \phi \\\ \end{matrix} \right|=\cos \theta \sin \theta {{\cos }^{2}}\phi +\cos \theta \sin \theta {{\sin }^{2}}\phi =\sin \theta \cos \theta$, taking $\sin \theta \cos \theta$ common from both the terms and putting ${{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1$ .
Multiplying the above determinant by $\cos \theta$,
$\cos \theta \left| \begin{matrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\\ -\sin \theta \sin \phi & \sin \theta \cos \phi \\\ \end{matrix} \right|=\sin \theta \cos \theta \times \cos \theta =\sin \theta {{\cos }^{2}}\theta$
By adding all the values we will get,
$\Delta ={{\sin }^{3}}\theta {{\cos }^{2}}\phi +{{\sin }^{3}}\theta {{\sin }^{2}}\phi +\sin \theta {{\cos }^{2}}\theta$
Take ${{\sin }^{3}}\theta$ common from the first two terms,
$\Rightarrow \Delta ={{\sin }^{3}}\theta ({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )+\sin \theta {{\cos }^{2}}\theta$
Put, ${{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1$
$\Rightarrow \Delta ={{\sin }^{3}}\theta +\sin \theta {{\cos }^{2}}\theta$
Take $\sin \theta$ common from both the terms,
$\Rightarrow \Delta =\sin \theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)=\sin \theta$
Hence the value of the determinant is $\sin \theta$.
Therefore the value is not a constant term. So option (c) is not correct.
It is free from $\phi$, that means the value of the determinant is independent of $\phi$.
Therefore, option (b) is correct.
The value of the determinant is $\sin \theta$. Therefore the value of the determinant is dependent on $\theta$.
Hence, option (a) is not correct.
Now to check option (d) let us differentiate the result with respect to $\theta$,
$\dfrac{d\Delta }{d\theta }=\dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta$
${{\left. \dfrac{d\Delta }{d\theta } \right|}_{\theta =\dfrac{\pi }{2}}}=\cos \left( \dfrac{\pi }{2} \right)=0$
Hence, option (d) is correct.
Therefore, option (b) and option (d) are correct.

Note: Find the value of the determinant very carefully. Take the negative and positive sign correctly while breaking the determinant.Use Trigonometric identities and formulas for simplification.